In this module we are going study about
•
Magnetic fields produced by currents
•
Ampere’s Law and Biot-Savart law
•
Motion of a charged particle in magnetic field
•
The force on a current in a magnetic field
•
The torque on a current carrying coil
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Galvanometer, Ammeter and Voltmeter
Electromagnetism-I
Success through conceptual learning
ELECTROMAGNETISM - I
Sources of magnetic field :1. The first magnetic phenomena observed were those associated with naturally occurring magnets fragments of iron ore found near the ancient city of magnesia (whence the term “magnet”). These natural magnets attract unmagnetized iron, the effect is most pronounced at certain regions of the magnet known as its poles.
2. In the year 1819, Danish scientist H.C. Oersted observed that a pivoted magnet (a compass needle) was deflected when in the neighborhood of a wire carrying current. Twelve years later Michael Faraday found that a momentary current existed in a circuit while the current in a nearly circuit was being started or stopped. Shortly afterward it was discovered that motion of magnet toward or away from the circuit would produce the same effect.
3. The work of Oersted demonstrated that magnetic effect could be produced by moving electric charges and that of Faraday and Henry showed that currents could be produced by moving magnets.
4. The magnetic field :
a) An electric charge sets up or creates an electric field E in the space surrounding it. b) The electric field E exerts a force F Eq = on a charge q placed in the field.
Similarly
a) A moving charge or a current sets up or creates a magnetic field in the space surrounding it. b) The magnetic field exerts a force on the moving charge or current carrying conductor in the
field.
5. A static charge produces only electric field but moving charge produces both electric field and magnetic field in space.
6. Moving charge current is the source of magnetic field. 7. Magnetic field of a moving charge :
The magnetic field due to moving charge q is 0 2 qv sin B 4 r µ θ = π .
Where v is velocity of charge, r is distance from charge, θ is angle between
v and r. 0 2 ˆ q v r B 4 r µ × = π
The direction of B is perpendicular to plane containing v and r.
8. Ampere established relationship between the current in the conductor and strength of the magnetic field around the conductor.
9. Oersted’s experiment : A magnetic compass needle placed in the vicinity of a conductor carrying conductor aligned perpendicular to the conductor.
10. The direction of deflection of north pole of magnetic needle is given by Ampere’s swimming rule. 11. Ampere’s swimming rule : Imagine that a man is swimming along the conductor in the direction
of the current facing a magnetic needle the north pole of the needle will deflect towards his left hand. S N i q v B r θ
12. The magnetic lines of force around a current carrying conductor are concentric circles with their center lying on the conductor.
13. The direction of magnetic field around the current carrying conductor is given by
a) Maxwell’s cork screw rule b) Right hand thumb rule
Maxwell’s cork screw rule : Imagine a right hand screw is advancing in the direction of current in a conductor. Then the direction of rotation of the screw gives the direction of magnetic lines of induction.
Right hand thumb rule : Imagine that a current carrying conductor is held in the right hand palm such that the direction of current is indicated by the thumb. Then the other fingers indicate the direction of magnetic lines of induction.
14. Ampere’s law : Ampere’s law states that the line integral of dlB. along a closed path round the current carrying conductor is equal to µ0i where i is the current through the surface bounded by the closed path and µ0 permeability of free space. i dl . B =µo
∫
i r 2 B π =µo r 2 i B o π µ =15. Biot-Savart’s law : The magnetic induction at a point near a current carrying conductor is directly proportional to the length of the conductor, the strength of the current and sine of the angle made by the line joining the point with the conductor and inversely proportional to the square of the distance of the point
o o 2 3 i.dl.sin i dB . ; dB . d lxr 4 r 4 r µ θ µ = = π π
The direction of magnetic induction is given by the vector (d i rc) × . This law is valid only for small current segments.
i
i r B
dl (b) Right hand thumb rule.
Thumb shows the direction of current and curled fingers show the direction
of lines of induction
↑ i B
i
(a) The direction of lines of induction around a current
carrying conductor (Right Hand Screw Rule)
dB C P r Q i θ d
16. Ampere's law and Biot-Savart law are equivalent but Ampere's law is more useful in some symmetrical conditions.
17. The magnetic induction at a point P due to a conductor of finite length is ) sin (sin r 4 i B 0 α+ β π µ =
18. The magnetic field induction due to the current carrying conductor of infinite length is given by r 2 i B or r i 2 4 B 0 0 π µ = ⋅ π µ =
Magnetic field at one end of infinite long conductor is B =
r 4 i 0 π µ
19. The magnetic field induction at a point along the axis of a circular coil is
2 / 3 2 2 2 0 ) x r ( nir 2 4 B + π ⋅ π µ =
where n = number of turns, i = current in the coil, r = radius and x = the distance of the point from the centre of the coil.
2 / 3 2 2 0 ) x r ( niA 2 4 B + ⋅ π µ = If x >> r, then 3 0 x niA 2 4 B ⋅ π µ = i.e., 3 0 x M 2 4 B ⋅ π µ =
where M= magnetic moment of the current loop.
20. (a) The magnetic field induction at the centre of a circular coil of n turns is given by relation
r ni 2 B = µ0 ⋅
(b) Magnetic field induction at the centre of an arc is B = 0i
4 r µ α
π
Ex-1 . An equilateral triangular loop of edge d carries a current i. Determine the magnetic field induction at its centroid.
Sol. The loop can be considered to be made up of three straight wires AB, BC and CA, each of length d and carrying a current i as shown in figure.
Let us first find the magnetic induction B1 due to the wire AB at the centroid O. Here, a = OM = 1(CM) 1dcos d 3 3 6 2 3 π ⎛ ⎞ = ⎜ ⎟⎝ ⎠= and θ1 = θ2 = 3 π.
Using equation B 0 i(sin 1 sin )2
4 a µ θ + θ = π , B = B 0 i sin sin 4 (d / 2 3) 3 3 µ ⎛ π π⎞ = ⎜ + ⎟ π ⎝ ⎠ = 0 6i 4 d µ
π directed into the plane of the page.
B r i A B r β α P i α i B x r i 2 2 r +x I A I I π/3π/3 π/6 O d B C
Similarly, the magnetic inductions at O due to the wires BC and CA are also 0 6i
4 d µ
π each directed along the inward normal to the plane of the page.
∴ The net magnetic field induction at O is B = 3B1 = 9 i0
2 d µ π .
Ex-2. For the wire PQRST shown in figure, find the magnetic induction B at the point O.
Sol. The entire wire can be divided into four parts as PQ, QR, RS and ST. Obviously, the magnetic fields due to each portion, at O is directed toward the inward normal to the plane of page. It remains for us to determine the field inductions B1, B2, B3 and B4 corresponding to the four parts respectively and add them to get the resultant field induction B.
Making use of equation B 0 i(1 sin )
4 a µ − θ = π B1 = B4 = 0 i 1 sin 4 (d / 2) 4 µ ⎡ − π⎤ ⎢ ⎥ π ⎣ ⎦
Making use of equation B 0 isin
4 a µ θ = π B2 = B3 = B 0 isin( / 4) 4 d µ π = π = 0 i 4 d 2 µ π ∴ B = 2(B1 + B2) = 0 2i 2 1 4 d 2 µ ⎡ + ⎤ ⎢ ⎥ π ⎣ ⎦ B = 3 i0 2 2 d µ π .
21. For a solenoid of finite length at any point on the axis B = [sin sin ]
2 Ni
0 α+ β
µ
N is number of turns per unit length.
22. A solenoid consists of closely wounded helical coil. Inside the solenoid the field is almost uniform. The magnetic induction at the centre of the solenoid is
l ni
B = µ0 where l is the axial length and n is total number of turns in length l of the solenoid. The equation holds good only when the radius of the turns is very small when compared with the length.
23. When current is passed through a helical spring, it contracts due to mutual attraction between consecutive turns.
24. Magnetic field at a point due to a cylindrical current carrying wire
Case (A) : When the wire is solid
Consider a solid cylindrical wire of radius R carrying a current I distributed uniformly across its cross-section. Let P be a point distant r from the axis, where the magnetic field induction B is to be determined. α β R L +∞ S i d i R Q P O i T – ∞ M Q P O – ∞ π/4d/√2 I d B O R P r
We consider a circular arc of radius r coaxial with the wire, such that the former passes through point P. From symmetry B must be tangential at each point on the curve and should also be equal in magnitude. Application of Ampere’s law
∫
B d ⋅ I= µ0i yields,B(2πr) = µ0 i
where i is the current embraced by the closed circular arc. The following two cases arise. i) When P lies within the cross-section of the wire, i.e., r < R :
Obviously, I I 2 2 2 r r i R R π ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ π ∴ B(2πr) = µ0I 2 r R ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ or, B = 0I 2 r i 2 R µ = π . …(i)
ii) When P lies outside the cross-section of the wire, i.e., r > R : Obviously, i in this case will be I.
∴ B(2πr) = µ0I or, B = 0I
2 r µ
π . …(ii)
It is evident that, at the surface of the wire r = R. Both equations (i) as well as (ii) yield consistent and expected results, viz., B = 0I
2 r µ
π
Figure displays graphically the variation of B with distance r from the axis.
Case (B) : When the wire is hollow and thin walled
As discussed above, with the same reasoning and using Ampere’s law it is easy to see that
B = 0 (within the wire) and B = 0I
2 r µ
π (outside the wire)
The variation of B and r is shown graphically in figure.
Ex-3. A straight wire of circular cross-section of radius R carries a total current I distributed in such a manner that j varies directly as distance form the axis. Find the magnetic field induction B at a distance r from the axis (where r < R).
Sol. Let P be the point where B is to be determined. We draw a circle of radius r coaxial with the wire so that P lies on it. Next, we consider a circular strip of infinitesimally
small thickness dr. Current flowing through this strip is dI = j(2πrdr) = (kr) (2πrdr) where k is a constant or, Total current I =
R 3 2 0 2 kR (k2 )r dr 3 π π =
∫
. ∴ k = 3I3 2 Rπ . 0I 2 R µ π B O R r R r 0I 2 R µ π B O R r R r R I Pd I d B O R P r drApplying Ampere’s law, yields
∫
B dl⋅ = µ0i i = r 3 3 0 3Ir r 2 dr I R 2 R ⎛ ⎞ π = ⎜ ⎟⎛ ⎞ ⎜ π ⎟ ⎝ ⎠ ⎝ ⎠∫
∴ B(2πr) = µ0 3 r I R ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ B = 0 2 3 r I 2 R µ π .25. The magnetic field induction due to the long current carrying cylindrical conductor is
2 0 ) r R ( ir 2 B + ⋅ π µ =
Where R is the radius of the conductor and r is the distance of the point from the surface of the conductor at which the value of B is given. However if r >> R, then
r 2 i B 0 π µ =
26. Force on a moving charge :
27. An electric charge moving in a uniform magnetic field experiences a force (F). F q(v x B)= or F qvB sin = θ
28. The direction of force is obtained by right hand grip rule. When the charge enters into a uniform magnetic field perpendicular to its direction, then F = qvB. If it enters along the direction of the field, F = 0.
29. The force can only change the direction in which the charge is moving but not its speed. Hence no work is done by it.
30. When a charged particle moves in a uniform magnetic field at right angles to the direction of the field.
i) the trajectory of motion changes and
ii) the speed and energy of a particle remain the same.
31. When a charged particle moves in an electric field, work is done and hence its kinetic energy increases.
32. If we assume that the earth’s magnetic field is due to a long circular loop of current in the interior of the earth, then the plane of the loop will be east-west and the current passes in clockwise direction when seen from earth’s north pole.
33. An electron is moving vertically downwards at a certain place. The direction of force on it due to the horizontal component of the earth’s magnetic field is towards west.
q v B Fmax = q v B +q B v
34. The force acting on a charged particle when it enters a uniform magnetic field of induction B, with velocity v at right angles to the field will provide the necessary centripetal force and make the charged particle move along a circular path of radius ‘r’.
F = qvB r mv2 = ; m qB r v or qB mv r= = Since T= 2π ω , we have 2 m T qB π =
Thus time period is independent of the particle-speed. Hence the faster particles move in bigger circles and slower particles move in smaller circles such that the period of revolution T is the same. 35. The frequency m 2 qB ) ( π =
ν and is called cyclotron frequency.
36. If a charged particle enters a uniform magnetic field along the line of the field, it goes undeviated. If it enters at an angle of inclination other than 90° to the field, its path is helical.
37. Helical motion of a charged particle in a uniform magnetic field : Consider a uniform magnetic field B oriented along
x-axis, and a charged particle having a charge q and mass m projected along the X-Y plane, making an angle θ with the X-axis. The component of velocity along the direction of B (i.e., v cos θ) does not contribute to the magnetic force, whereas the component, normal to the direction of B(i.e., v sin θ), contributes for the magnetic force. Thus,
F = qvB sin θ
This provides the necessary centripetal force, for rotational motion, whereas the component of velocity v
cos θ, possessed by the particle imparts a movement along the magnetic field. The overall motion is said to be helical, wherein, the particle moves along a spring shaped path (figure).
Notice, that the projection of the path of the particle, in a plane normal to the magnetic field, is a circle (Y-Z plane in the diagram).
The particle rotates about the magnetic field (axis of the helix) with a period of revolution T given by, T 2 m
qB π =
which is independent of the speed.
The particle traverses along the magnetic field with a velocity v cos θ; the distance traveled along the helical path axis, in one period of revolution (T) is known as the pitch of the helix (p in the diagram) which is given by
2 mv cos p
qB
π θ
= .
It should be noted that the particle started off from the X-Z plane and touches time and again, after each period of revolution (T). Thus, it can be concluded that, the charged particle touches the plane, from where it started, at times t = nT, where n ∈ I.
v +q R B O F F F F v v v F v p v θ +q B v cosθ X Z Y v sin θ B
Ex-4. A magnetic field of induction B B i = 0ˆ exists along the X-axis. A particle of charge q and mass m is projected from the origin O along the X-Y plane at an angle α to the X-axis with a speed v. (see figure). Find the position coordinates (x, y, z) and the velocity v of the particle after time t of the projection.
Sol. The initial velocity v of the particle can be expressed as
0 ˆ ˆ
v =v(cos i sin j)α + α .
The component v cos iαˆ along the magnetic field remains constant as no force acts along B
(ie., X-axis).
Therefore x = (v cos α) t.
Now, the projection of the helical path of the particle on the Y-Z plane (normal to B) is a circle as shown in figure. The lorentz force (magnetic force) providing the necessary centripetal force is
2 m(v sin ) qvBsin R α α =
where R is the radius of the helix. or R mv sin
qB α
= .
Also, the angular speed ω for the rotation of the particle about the pitch axis is
2 2 (v sin ) qB
T 2 R m
π π α
ω = = =
π .
∴ The angle rotated about the pitch axis (an imaginary line passing through the centre C and parallel to the X-axis) in time t, is
t qBt
m θ = ω =
Figure shows the projection of the particle’s path; P is the projection of particle at any instant.
The y coordinate is y = PM = R sin θ = mv sin sin qBt
qB m
α ⎛ ⎞
⎜ ⎟
⎝ ⎠.
where the sign of sin(qBt/m) automatically takes account all positions of the particle. The z coordinate is z = –OM = –(OC – MC) i.e., z = R(1 – cos θ) = mv sin 1 cos qBt qB m ⎡ ⎤ α ⎛ ⎞ − ⎢⎣ − ⎜⎝ ⎟⎠⎥⎦.
Thus, x = v cos αt, y = mv sin sin qBt
qB m
α ⎛ ⎞
⎜ ⎟
⎝ ⎠
and z = mv sin 1 cos qBt
qB m ⎡ ⎤ α ⎛ ⎞ − ⎢ − ⎜ ⎟⎥ ⎝ ⎠ ⎣ ⎦.
Coming to the velocity of the particle, the component parallel to X-axis remains unchanged.
∴ vx = v cos α.
From figure, the velocity normal to the X-axis, (i.e., in the Y-Z plane) remains constant in magnitude, but has rotated by an angle θ from its initial direction (positive Y-axis).
R M X v sin α –Z Y C O P θ v sinα –Z Y v sinα sinθ θ v sin α cos θ v q X Y α Z B X Y Z O R θ C F F
∴ vy = v sin α cos θ = v sin α cos qBt m
⎛ ⎞
⎜ ⎟
⎝ ⎠
and vz = – v sin α sin θ = –v sin α sin qBt m
⎛ ⎞
⎜ ⎟
⎝ ⎠
∴ v=v i v j v kxˆ+ yˆ+ zˆ
or, v v cos i sin cosˆ qBt ˆj sin sin qBt kˆ
m m ⎧ ⎛ ⎞ ⎛ ⎞ ⎫ = ⎨ α + α ⎜ ⎟ − α ⎜ ⎟ ⎬ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ .
38. Motion of a charged particle in space containing both electric and magnetic fields :
If a region contains electric and magnetic fields E and B respectively and a particle having a charge q is in motion with a velocity v, then the force acting on it is given by
F q(E v B)= + × known as Lorentz force.
Several cases are possible, depending upon the situation and complexity. Here, we intend to discuss two simple but important (from particle point of view) cases viz, (i) E is parallel to B and (ii) E is perpendicular to B.
In each case the initial velocity v0 of particle will be taken normal to both E and B. (i) E is parallel to B : Consider a situation in which a constant electric field
E = E i0ˆ and a constant magnetic field B B i = 0ˆ exist in space pointing along the X-axis. Let a particle having a charge q and mass m be projected with an initial velocity u the along Y-axis. Let us analyse its further motion.
The case is too simple in the sense that the speed of the particle normal
to the direction of E, (i.e., the velocity u ) remains constant as no work is done on or against it in the direction. It follows that the particle executes rotatory motion about an axis parallel to the X-axis as a helix due to the magnetic force. However, the electric force qE keeps on accelerating the particle which as a result, circulates the helical path once in the same time, but the increasing speed amounts to the pitch of the helix, getting continuously
enhanced with time. Thus, the acceleration along X-axis due to electric force is 0 x x qE F a m m = = .
∴ The velocity vx at time t will be vx = axt = qE t0
m .
The initial velocity u (normal to the field) makes the charged particle rotate about the X-axis, as shown in figure. Applying RHR-1 to the particle at origin (initially), it is obvious that the centripetal force (magnetic force) acts along the negative Z-axis. Hence, the projection the helical path on the Y-Z plane will be as shown in the figure. Thus, the velocity components and position coordinates corresponding to the Y-Z plane will be identical to that discussed in Example, with the only substitution of v = u and α = π/2, so that
sin α = 1 and cos α = 0. y = musin qBt qB m ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ z = mu 1 cos qBt qB m ⎡ ⎛ ⎞⎤ − ⎢⎣ − ⎜⎝ ⎟⎠⎥⎦ u –Z Y C O P θ Z u E X B Y C u X Z Y O
vy = u cos θ = u cos qBt m
⎛ ⎞
⎜ ⎟
⎝ ⎠
and vz = – u sin θ = – u sin qBt m ⎛ ⎞ ⎜ ⎟ ⎝ ⎠. Using s = ut 1at2 2
+ along X-axis, where E produces a constant acceleration, the displacement of the particle along X-axis is
x = 0 1qE0t2
2 m
⎛ ⎞
+⎜ ⎟
⎝ ⎠.
∴ The instantaneous position vector of the charged particle can be expressed as : r=xi yj zkˆ+ +ˆ ˆ 2 0 qE t ˆ mu qBt ˆ mu qBt ˆ r i sin j 1 cos k 2m qB m qb m ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ =⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ − ⎜⎝ − ⎟⎠ 0 qE tˆ qBt ˆ qBt ˆ v i ucos j usin k m m m ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ − ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ .
(ii) E is perpendicular to B : We know venture into the case where E E i = 0ˆ and B B j = 0ˆ, and a charged particle is released at the origin O. The electric field keeps on accelerating it along the X-axis, while the magnetic field B (directed along Y-axis) makes the particle rotate in a plane normal to it, (i.e., the X-Z plane). There is no movement along the Y-axis, as neither there is any initial velocity component nor there is any acceleration in that direction. Thus, the overall motion is purely in the X-Z plane.
First of all let us express the velocity at any instant as
xˆ zˆ
v=v i v k+
where vx and vz are the components of the velocity along X and Z axis respectively. Net force on the particle at time t is
e m F F = +F qE q(v B) = + × 0ˆ xˆ zˆ 0ˆ q[E i (v i v k)B j] = + + , i.e., F q(E= 0−v B )i qv B k F i F kz 0 ˆ+ x 0ˆ= xˆ+ zˆ so that Fx = q(E0 – vz B0) and Fz = qvxB0.
∴ The acceleration along X and Z axis will be respectively
x x 0 z 0 F q a (E v B ) m m = = − and x z 0 z qv B F a m m = = . Now, x 0 z 0 z qB qB da dv (a ) dt m dt m − ⎛ ⎞ = ⎜⎝ ⎟⎠= − = qB0 qv Bx 0 m m ⎛ ⎞ − ⎜⎝ ⎟⎠ or, 2 2 0 x x 2 qB d v v 0 m dt ⎛ ⎞ +⎜ ⎟ = ⎝ ⎠ .
This is the differential equation of a SHM whose general solution can be expressed as vx = v1 sin (ωt + δ)
where ω2 = qB0 2
m
⎛ ⎞
⎜ ⎟
⎝ ⎠ and ; v1 and δ are constants which depend upon a particular situation and can be obtained by the use of boundary conditions.
Initially, at time t = 0, vx = vy = vz = 0. ∴ 0 = v1 sin (δ) or, δ = 0. ∴ vx = v1 sin ωt Also, x 1 x dv v cos( t) a dt = ω ω = .
Again, initially at time t = 0, ax = qE0 m . ∴ qE0 m = ωv1 or, v1 = 0 qE mω . But, ω = qB0 m ∴ v1 = 0 0 E B . ∴ vx = 0 0 E B sin ωt …(i) Again, from az = z x 0 dv qv B dt =m . Substituting for vx, we have
az = qE0sin t dvz m ω = dt .
Integrating between the limits t = 0, vz = 0 and t = t and vz = vz, vz = qE0(1 cos t)
mω − ω …(ii)
Integration of expressions corresponding to vx and vz between the limits, t = 0, vx = 0, vz = 0 and t = t, vx = vx, vz = vz, yields x = 0 0 E (1 cos t) B ω − ω …(iii) and z = 0 0 E ( t sin t) B ω ω − ω …(iv) where ω = qB0 m . …(v)
Thus, equation (i) and (ii) combinedly give the velocity at any instant while equations (iii) and (iv) yield the position coordinates on the X-Z plane.
The path traveled by the particle is a cycloid. In general, a cycloid is represented by the parametric form :
x = a(θ ± sinθ) and y = a(1 ± cosθ)
where ‘a’ is known as the radius of the generating circle, and 2πa is the base of the cycloid. In our discussion see figure,
a = 0
0
E B ω.
39. The frequency of a charged particle in a uniform magnetic field is known as cyclotron frequency as the particles in a circular accelerator, called cyclotron, move with this frequency.
40. Cyclotron is a device used to accelerate charged particles to high speed for nuclear reaction. It was invented by Lawrence.
2a E C D A –Z O X 2πa 2πa
41. Force on a current carrying conductor :
42. When a conductor of length l carrying a current i is placed at an angle θ to the direction of the magnetic field of flux density or induction B, the force on the conductor or wire is given by
F = (il xB)orF=ilBsinθ
Its direction can be determined by using Fleming’s left hand rule, whose statement is as follows:
Stretch the fore finger, middle finger and thumb of the left hand mutually perpendicular to each other. If the fore finger represents the direction of magnetic field, the middle finger that of current, then the thumb will represent the direction of force on the conductor.
Ex-5. Figure shows a circular arc ab of radius r carrying a current i resting in a uniform magnetic field of strength B. Find the magnitude and direction of the force (magnetic) exerted on the arc.
Sol. The length of straight line joining ab is
2r sin 2 θ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ .
The required force is Fnet =i(ab B)× or, Fnet 2Bir sin
2 θ ⎛ ⎞
= ⎜ ⎟
⎝ ⎠.
(in a direction normal to the line joining ab to the right). 43. Force between two current carrying conductors :
If i1 and i2 are the strengths of currents passing through two infinitely long, straight and parallel wires separated by a distance r, the magnetic field induction B, due to the flow of current in the first conductor at a distance r on the second conductor is
r 2 i B1 01 π µ = . The force exerted by this induction on a length l on the second conductor is
F =i2B1l
The force per unit length is
r 2 i i F 012 π µ = .
If the current is in the same direction, there will be attraction and if the current is passing in opposite direction, there will be repulsion.
44. An ampere is that steady current when flowing in each of two long straight parallel wires separated by a distance of one metre apart in vacuum causes each wire exert to each other a force of 2 × 10–7 N per each metre length of wire.
F
B
v or
i
Fleming’s left hand rule
B i l i i F B i B i a r b θ r i1 i2 r
Ex-6. A square loop of side a carries a current i2 with a long straight conductor carrying a current i1 lying in the plane of the loop, with its closer edge, which is parallel to the conductor, distant b. Find the force exerted on the loop.
Sol. The relevant diagram is shown in figure. The magnetic field due to the conductor is normal to the plane of the square loop. The total force on the loop can be obtained as the vector sum of the forces on its four arms, namely pq, qr, rs and sp respectively.
Let the current i2 flow in the loop in the clockwise sense as shown by arrows.
Evidently, the relative configurations of the edges qr and ps with respect to the long conductor are identical and hence the forces on
them will be equal in magnitude; however, the current directions in them being opposite, the forces cancel each other.
Force on side pq is given by F1 = B1i2 a(sin 90°)
where B1 is the magnetic field at the site of pq, which is equal to 0 1i 2 a µ π . ∴ F1 = 0 1 2i i 2 µ π . (toward left) And, force on the side rs is
F2 = B2i2 a sin 90°
where B2 is the magnetic field at the location of rs, which is equal to F2 = 0 1i 2 (a b) µ π + ∴ F2 = 0 1 2i i a 2 (a b) µ π + (toward right)
∴ The net force (toward left) acting on the loop is
Fnet = F1 – F2 = 0 1 2i i b
2 (a b) µ
π + .
45. Force and torque on current loop or coil in a magnetic field :
The total magnetic force on any closed current loop in a uniform magnetic field is zero. 46. Torque or couple on a current carrying rectangular conductor or
loop in a uniform magnetic field is given by τ = iAB sinθ where A is the area of the rectangular loop, B is the magnetic induction and θ is the angle between the normal to the plane of the rectangular loop and the field. If there are n turns, then the torque acting on the coil suspended is given by τ = ni AB sin θ or τ = MB sinθ where M is the magnetic dipole moment equal to niA and θ is the angle between
the magnetic induction and the normal to the plane of the coil i.e., the direction of the magnetic moment. When the plane of the coil makes an angle θ with the field, then the couple acting on the coil is
τ = niAB Cosθ or τ = MB cos θ.
i P B Q R S i Bil Bil θ i i p s i1 F2 r q –∞ +∞ ⊗ ⊗ i2 F1 b a
Moving coil galvanometer :
47. A moving coil galvanometer consists of a powerful horse shoe magnet with concave poles to produce a uniform radial magnetic field. On a cylindrical soft iron core, a coil is wound and the coil is suspended with a phosphorbronze fibre. The plane of the coil always lies in the direction of the magnetic field because the magnetic field is radial. The whole apparatus is kept inside a brass case provided with a glass window.
NAB C
i= θ ampere
Here C is the couple per unit twist on the suspension fibre, ‘i’ is the current passing through the galvanometer coil and θ is the angle of deflection by the needle.
C/BAN is a constant (k) is called as galvanometer constant. Deflection is measured more accurately using Lamp and Scale arrangement. The ray reflected onto the scale by the mirror is deflected by 'x'. The distance between the mirror attached to the suspension fibre of the galvanometer and the scale is D, then
Tan 2θ ≈ 2θ when 2θ is very small Then 2D x ; D x 2θ= θ= ; i = 2D x . BAN C
48. The current sensitivity of a galvanometer is the deflection in mm produced on a scale kept at a distance of one metre by a constant current of one microampere.
49. The reciprocal of the current sensitivity is called figure of merit and is expressed in µA/mm. 50. A current upto 10–9A can be measured using moving coil galvanometer.
51. Table Galvanometer : It has a rectangular coil of insulated copper supported on two bearings. The poles of the magnet are concave. It has a light aluminium pointer which moves on a scale. The whole arrangement is kept in an ebonite case with a glass window.
52. Tangent Galvanometer :
a) It is portable and minimum measurable current is of the order of 10–6 A.
b) works on the principle of Tangent law B = BHTanθ
Here B = Magnetic induction due to passage of current in the coil = r 2 i 0 µ
c) current measured by Tangent galvanometer is
⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ µ = n rB 2 i 0 H Tanθ = KTanθ r = Radius of coil
n = number of turns of coil
d) Reading is more accurate when θ = 45° since relative error di 1
i ∝sin2θ and it is minimum for 45°
e) Sensitiveness is maximum when θ = 0° since sensitiveness
di
dθ∝ cos2θ, which is maximum
for θ = 0°.
f) During experiment, plane of the coil should be along the magnetic meridian [to fulfill the requirement of tangent law]
i P B Q R S b Bil Bil θ i
θ i B H B
53. Shunt :
a) A low resistance connected in parallel to galvanometer to protect it from large current is known as shunt. b) i = ig + is c)
( )
S G S i ig + = d)( )
S G G i is + = e) n 1 S G S i i s g = + =f) effective resistance of circuit = S G
GS +
g) If the range of galvanometer is increased to n times, 1/nth of main current passes through galvanometer. Hence sensitiveness decrease by n times.
54. Ammeter :
a) It is a device used to measure current in electrical circuits.
b) Galvanometer can be converted in to Ammeter by connecting low resistance parallel to it. c) To increase the range by ‘n’ times or to decrease the Sensitiveness by ‘n’ times, shunt to be
connected across Galvanometer.
( )
(
n 1)
G ig i G ig S − = − = Here n = A / divisions new A / s dividision old range old range new i i g = = d) Resistance of Ammeter = S G GS +e) Resistance of ideal Ammeter is zero and its conductivity is infinity f) Ammeter must be always connected in series to the conductor.
g) Among low range and high range Ammeter, low range Ammeter has more resistance. h) As shunt value decreases sensitivity decreases, accuracy increases.
55. Voltmeter :
a) Voltmeter is a device used to measure P.D. across the conductor in electric circuits.
b) Galvanometer is converted into voltmeter by connecting high resistance in series to it.
c) Voltmeter is always connected in parallel to the conductor [P.D. across which is to be measured]
d) Resistance of voltmeter = G +R =
ig V
e) Here ‘V’ is range of voltmeter (e) resistance of ideal voltmeter is infinity and conductivity is zero. f) Among low range and high range voltmeters, high range voltmeter has more resistance. g) P.D. across the ends of voltmeter is, V = ig + (G + R)
h) Resistance to be connected in series to galvanometer to convert into voltmeter is G i V R
g −
= .
i) To increase the range by n times, n 1 2 rangeV old rangeV new =
(
)
( )
g g i G R R 1 i G G + = ⇒ +Hence resistance to be connected in series to galvanometer is R = G(n – 1). j) As series resister value increases sensitivity decreases, accuracy increases.
i – ig S i ig G i – ig R r i ig G
Assertion & Reason : In each of the following questions, a statement is given and a corresponding statement or reason is given just below it. In the statements, marks the correct answer as
1) If both Assertion and Reason are true and Reason is correct explanation of Assertion. 2) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. 3) If Assertion is true but Reason is false.
4) If both Assertion and Reason are false. 5) If Assertion is false but Reason is true.
1. [A] : Magnetic field interacts with a moving charge and not with stationary charge. [R] : A moving charge produces a magnetic field.
2. [A] : A linear solenoid carrying current is equivalent to bar magnet. [R] : The magnetic field lines of both are same.
3. [A] : Force experienced by moving charge will be maximum if direction of velocity of charge is perpendicular to applied magnetic field.
[R] : Force on moving charge is independent of direction of applied magnetic field. 4. [A] : Torque on coil is maximum, when coil is suspended in radial magnetic field.
[R] : Torque depends upon the magnitude of the applied magnetic field. 5. [A] : The coil is wound over the metallic frame in moving coil galvanometer.
[R] : The metallic frame help in making steady deflection without any oscillation.
6. [A] : When two long parallel wires, hanging freely are connected in series to a battery, they come closer to each other.
[R] : Wires carrying current in opposite direction repel each other.
7. [A] : When the observation point lies along the length of current element, magnetic field is zero.
[R] : Magnetic field close to current element is zero.
8. [A] : A moving coil galvanometer has a permanent magnet with cylindrical pole faces.
[R] : The radial magnetic field produced keeps the constant of the galvanometer to be constant throughout the deflection range.
9. [A] : A moving coil galvanometer has a permanent magnet with cylindrical pole faces.
[R] : The radial magnetic field produced makes the coil to experience a constant torque in any deflected position.
10. [A] : A moving coil galvanometer has a permanent magnet with cylindrical pole faces.
[R] : The radial magnetic field produced makes the calibration of the scale to be uniform throughout the scale.
11. [A] : A soft iron cylinder is arrange in the space of the coil suspended between the pole pieces of the magnet.
[R] : To increase the sensitivity of the galvanometer. 12. [A] : The resistance of a voltmeter is high.
[R] : The voltmeter is connected always in parallel. 13. [A] : The resistance of an ideal voltmeter is infinite.
[R] : When connected the voltmeter should not draw any current from main circuit thus keeping the current in that circuit constant.
14. [A] : The resistance of potentiometer during the measurement of potential differences, can be considered as infinite.
[R] : It does not draw any current from secondary circuit. 15. [A] : The resistance of an ammeter is very low.
[R] : The ammeter is always connected in series. 16. [A] : The ideal ammeter has zero resistance.
[R] : The ammeter is always connected in series. As such it should not alter the current in the circuit.
17. [A] : The magnetic field ‘B’ due to a conductor of radius ‘r’ carrying a current ‘i’, at a distance of ‘x’ from the axis of the wire (x<r) is proportional to x.
[R] : The current is distributed uniformly along the area of cross-section of the wire. 18. [A] : An electron moves along the line AB which lies in the same plane as a
circular loop of conducting wire as shown in the figure current induces in the coil.
[R] : Moving electron produces a magnetic field. As it moves the magnetic field at a point changes.
19. [A] : When a current passing through a coil decreases, back emf is induced in it. [R] : As the current is decreasing the magnetic field produced in the coil decreases.
20. [A] : A Straight and very long wire carries current i. If the wire is vertically placed and carries current in the upward direction, null point will be formed on the west side.
[R] : If the wire is horizontal and carries current from west to east, null point will be formed at a point above the conductor.
21. [A] : A vertical straight wire carries a current vertically upwards. Points P and Q lie respectively to the east and west of the conductor at the same distance from it. At P the induction is BP, at Q the induction is BQ, then BP > BQ.
[R] : Two infinitely long thin insulated straight wires carrying currents i and i lie along x and y axes respectively. The locus of the points where the magnetic field is zero is straight line.
22. [A] : Two very long straight wires are connected in series with a battery and are arranged parallel to each other. Then the mutual force between the wires is repulsive.
[R] : If the two wires are connected in parallel with the battery and one arranged parallel to each other, the mutual force between wires is attractive.
23. [A] : Moving electric charge produce both electric and magnetic field.
[R] : A current carrying conductor produces a magnetic field in the surrounding space.
24. [A] : If a straight conductor carries current normally outwards from the plane of paper, lines of induction are concentric circles in counter clockwise direction.
[R] : A Straight conductor carrying current produces a magnetic field, which is radially symmetric. 25. [A] : An infinite long conductor carries a current i. The work done to move a unit north pole
round it once is µoi.
[R] : A straight cylindrical pipe carries current i, the field inside is zero.
26. [A] : For a given length l of a wire carrying a current i, the number of circular turns which produce the maximum magnetic moment is one.
[R] : A current carrying circular coil behaves as a magnetic shell.
27. [A] : When a charged particle moves at right angles to the magnetic field, the momentum of the particle will remain constant.
[R] : When a charged particle is moving in magnetic field, work done by the magnetic field on the charged particle is zero.
28. [A] : When a charged particle enters perpendicular to the direction of a uniform magnetic field in the time period of rotation depends on nature the particle.
[R] : When the particle enters the magnetic field at angle θ (≠0, ≠1800, ≠900), the path of the particle will be helical.
29. [A] : The force experienced by a semi circular wire radius r when it is carrying a current i and is placed in a uniform magnetic field of induction B as shown in fig(a) is zero.
[R] : The force experienced by a semi circular wire of radius r when it is carrying a current i and is placed in a uniform magnetic field of induction B as shown in fig(b) is 2B ir.
B (b) i (a) B i A e– B
30. [A] : If an electron is not deflected in passing through a certain region of space, then there is no electric and magnetic fields.
[R] : When a charged particle moves parallel to electric field its kinetic energy increases.
31. [A] : When a current carrying coil is placed in a uniform magnetic field, the net force on it always is zero.
[R] : The current carrying coil acts like a dipole.
32. [A] : In moving coil Galvanometer, phosphorous bronze wire is used as suspension wire. [R] : Phosphorous bronze has high young’s modulus , less rigidity modules.
33. [A] : For more current sensitivity, of Moving coil Galvanometer, B, A, N should be more and c should be small.
[R] : In moving coil Galvanometer, voltage sensitivity also varies as that of current sensitivity. 34. [A] : In Moving coil Galvanometer, concave shape magnetic poles makes current as linear
function of deflection(θ).
[R] : In case of redial magnetic field normal to the plane of coil always makes 90° with magnetic field.
35. [A] : In Moving coil Galvanometer external magnetic fields have no effect on the deflection. [R] : In tangent galvanometer external magnetic fields may influence the deflection.
36. [A] : Shunt increases the range of galvanometer. [R] : Shunt increases the life of galvanometer. 37. [A] : Ammeter must be always connected in series.
[R] : Conductance of ideal ammeter is infinity. 38. [A] : Voltmeter must be always connected in parallel.
[R] : Conductance of ideal voltmeter is infinity.
39. [A] : The resistance of ammeter is less than that of galvanometer.
[R] : In ammeter zero division is at one end of the scale, in Galvanometer zero division is at the middle of the scale.
40. [A] : A beam of charged particles is passed through a magnetic field. The work done on the beam by the field is dependent on the deflection of the beam.
[R] : The time rate of the work done by a magnetic field Bon a charged particle of moving in a helical path is zero.
41. [A] : On increasing length of the, Potentiometer wire, its sensitivity increases. [R] : On increasing length of the Potentiometer wire, potential gradient decreases. 42. To convert galvanometer into an ammeter, a shunt is used.
[A] : The accuracy of the meter increases. [R] : The range of the meter increases.
43. To convert a galvanometer into a voltmeter, a large resistance is connected in series with the galvanometer. Due to use of large resistance.
[A] : Accuracy of the voltmeter increases. [R] : Sensitivity decreases.
44. [A] : Range and least count of ammeter and voltmeter are linearly related with each other. [R] : Least count and sensitivity of an ammeter are related inversely with each other. 45. A current i is passing through a circular coil.
[A] : To induce magnetic induction of constant magnitude (at centre of the coil), the current required in it is directly proportional to its radius.
[R] : Induction of induced magnetic field at centre is inversely proportional to radius of the coil. 46. [A] : In Meter bridge, meter bridge wire is replaced by another wire having twice the cross
sectional area, the accuracy increases.
47. [A] : If a moving charged particle traces a helical path in a uniform magnetic field, axis of the helix is parallel to the magnetic field.
[R] : If in a region a uniform magnetic field and a uniform electric field both exist, a charged particle moving in this region cannot trace circular path.
48. [A] : In tangent Galvanometer plane of the current carrying coil must be in magnetic meridian. [R] : When plane of the coil is in magnetic meridian, B and BH are perpendicular to each other. 49. [A] : M.C.G can be used for the measurement of currents even in mines.
[R] : In case of M.C.G galvanometer constant does not depend on earth’s magnetic field. 50. [A] : A linear solenoid carrying current is equivalent to a bar magnet.
[R] : The magnetic field lines of both are same.
51. [A] : Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it.
[R] : The average velocity of free electron is zero.
52. [A] : Torque on the coil is maximum, when coil is suspended in a radial magnetic field. [R] : Torque depends upon the magnitude of the applied magnetic field.
53. [A] : The coil is wound over the metallic frame in moving coil galvanometer. [R] : The metallic frame help in making steady deflection without any oscillation.
54. [A] : Out of galvanometer, ammeter and voltmeter, resistance of ammeter is lowest and resistance of voltmeter is highest.
[R] : An ammeter is connected in series and a voltmeter is connected in parallel, in a circuit. 55. [A] : For a point on the axis of a circular coil carrying current, magnetic field is maximum at the
centre of the coil.
[R] : Magnetic field is inversely proportional to the distance of point from the circular coil.
56. [A] : A circular loop carrying current lies in XY plane with its centre at origin having a magnetic flux in negative Z-axis.
[R] : Magnetic flux direction is independent of the direction of current in conductor. 57. [A] : The energy of charged particle moving in a uniform magnetic field does not change.
[R] : Work done by the magnetic field on the charge is zero.
58. [A] : The electron passing through crossed magnetic and electric field is always deflected from its path.
[R] : If velocity of electron is equal to ratio of electric and magnetic field applied than electron beam remains undeflected.
59. [A] : A small coil carrying current, in equilibrium, is perpendicular to the direction of the uniform magnetic field.
[R] : Torque is maximum when plane of coil and direction of the magnetic fields is parallel to each other.
60. [A] : Earth’s magnetic field does not affect the working of a moving coil galvanometer.
[R] : The earth’s magnetic field is quite weak as compared to magnetic field produced in the moving coil galvanometer.
61. [A] : Soft iron core is used in a moving coil galvanometer. [R] : Soft iron is ferromagnetic in nature.
62. [A] : In ammeter, current in shunt is always greater than current in galvanometer.
[R] : Value of shunt resistance is negative if current in shunt is less than current in galvanometer. 63. [A] : No net force acts on a rectangular coil carrying a steady current when suspended freely in a
uniform magnetic field.
[R] : Forces acting on each pair of the opposite sides of the coil are equal and opposite.
64. [A] : If an electron and proton enter in an electric field with equal energy, then path of electron more curved than that of proton.
65. [A] : For a given charged particle moving in a given magnetic field, the radius of circular path is directly proportional to the momentum of particle.
[R] : The effect of magnetic field on a charge particle, change only its path from linear to circular. 66. [A] : If a proton and an α-particle enter a uniform magnetic field perpendicularly, with the same
speed, the time period of revolution of α - particle is double that of proton.
[R] : In a magnetic field, the time period of revolution of a charged particle is directly proportional to the mass of the particle and is inversely proportional to charge of particle.
67. [A] : If an electron while coming vertically from outerspace enter the earth’s magnetic field, it is deflected towards west.
[R] : Electron has negative charge.
68. [A] : The workdone by magnetic field on a moving charge is zero. [R] : In magnetic field force is perpendicular to the velocity.
69. [A] : An electron and proton enters a magnetic field with equal velocities, then the force experienced by proton will be more than electron.
[R] : The mass of proton is 1837 times more than the mass of electron.
70. [A] : A charged particle moving in a uniform magnetic field penetrates a layer of lead and there by loses half of its kinetic energy. The radius of curvature of its path is now reduced to half of its initial value.
[R] : Kinetic energy is inversely proportional to radius of curvature. 71. [A] : Tangent Galvanometer can not be used in mines.
[R] : In case of Tangent Galvanometer reduction factor depends on earth’s magnetic field. 72. Match the following :
List – I List – II
a) In MCG e) i α θ
b) in tangent galvanometer f) i α tanθ
c) Biot Savart’s law g) B= o 2
r sin dl i 4 θ π µ d) Ampere’s law h) B= r i 4 o π µ
1) a-e, b-f, c-g, d-h 2) a-h, b-g, c-f, d-e 3) a-f, b-e, c-h, d-g 4) a-g, b-h, c-e, d-f 73. Match the following :
List – I List – II
a) infinite resistance e) ideal voltmeter
b) zero resistance f) ideal ammeter
c) Lorentz’s magnetic force g) q(VxB) d) at centre of current
carrying circular loop h) B=
r i 2
o
µ
1) a-e, b-f, c-g, d-h 2) a-h, b-g, c-f, d-e 3) a-f, b-e, c-h, d-g 4) a-g, b-h, c-e, d-f 74. Match the following.
List – I List – II
a) right hand thumb rule e) magnitude of magnetic field b) Biot-Savart law f) direction of induced current c) Fleming’s left hand rule g) direction of magnetic field
d) Fleming’s right hand rule g) direction of force due to magnetic fields The correct code is
75. The magnetic inductions due to a circular coil carrying current at its centre, at distances on its axis x = R and x = 2R are represented by BC, BR, B2R. What are the ascending order of these values.
1) BC, BR, B2R 2) B2RBR, BC 3) BC, B2R, BR 4) BR, B2R, BC
76. The resistance of moving coil galvanometer is ‘G’ and that of an ammeter is RA and that of voltmeter is RV arrange them in decreasing order
1) G > RA > RV 2) RV > G > RA 3) G > RV > RA 4) RA > G > RV KEY : 1) 1 2) 1 3) 3 4) 2 5) 1 6) 5 7) 2 8) 1 9) 1 10) 1 11) 1 12) 2 13) 1 14) 1 15) 2 16) 1 17) 1 18) 1 19) 1 20) 2 21) 2 22) 2 23) 2 24) 1 25) 2 26) 2 27) 5 28) 2 29) 2 30) 4 31) 1 32) 1 33) 2 34) 1 35) 2 36) 2 37) 2 38) 3 39) 2 40) 5 41) 1 42) 2 43) 2 44) 2 45) 2 46) 5 47) 2 48) 1 49) 1 50) 1 51) 1 52) 2 53) 1 54) 2 55) 1 56) 3 57) 1 58) 5 59) 2 60) 1 61) 1 62) 1 63) 1 64) 4 65) 2 66) 1 67) 1 68) 1 69) 5 70) 4 71) 1 72) 1 73) 1 74) 1 75) 2 76) 2
MULTI CORRECT ANSWERS :
1. Choose the correct statements :
a) The work of Oersted demonstrated that magnetic effect could be produced by moving electric charge.
b) The work of Faraday and Henry demonstrated that magnetic effects could be produced by moving electric charge.
c) The work of Oersted demonstrated that currents could be produced by moving magnets d) The work of Faraday and Henry demonstrated that currents could be produced by moving
magnets.
2. Choose the correct statements :
a) The electric force on a charge does not depend on the speed of the charge b) The electric force on a charge depends on the speed of the charge.
c) The magnetic force on a charge does not depend on the speed of the charge d) The magnetic force on a charge depends on the speed of the charge.
3. F = qvBsinθ is the equation of the Lorentz force F on a charge q moving with a velocity v in a magnetic field B. Which of the following pairs of quantities are always perpendicular to each other?
a) F – v b) F – B c) v – B d) v – q
4. In which of the following situations, the magnetic field will not exert a force ? a) On a static charge, in the same direction as the direction of the magnetic field b) On a static charge, in a direction perpendicular to the direction of the magnetic field c) On a moving charge, in the same direction as the direction of the magnetic field d) On a moving charge, in a direction perpendicular to the direction of the magnetic field 5. Which of the particles passing normally through a magnetic field will not be affected ?
a) Electrons b) Neutrons c) Protons d) Neutrinos
6. A proton is moving in a uniform magnetic field. The initial direction of proton makes an angle θ with the magnetic field. Which of the following pairs representing θ and the path of the proton are correct ?
7. Both the electric an the magnetic fields can deflect an electron. Choose the correct alternatives. The force exerted by the
a) magnetic field changes the kinetic energy of the electron
b) magnetic field does not change the kinetic energy of the electron c) electric field changes the kinetic energy of the electron
d) electric field does not change the kinetic energy of the electron
8. Which of the following pairs representing the field and the possible path of the deflection of an electron beam in the field are correct ?
a) Uniform magnetic field – Circular. b) Uniform magnetic field – Helical c) Uniform magnetic field – Helical. d) Uniform electric field – Parabola
9. A particle (electron or proton) is moving vertically downward through a magnetic field directed from south to north in a horizontal plane. Which of the following pairs representing particle and the direction of deflection of the particle are correct ?
a) Electron – West b) Electron – East c) Proton – West d) Proton – East 10. Choose the correct statements :
a) The magnetic lines of force always emanate from the north pole, follow a curved path and end at the south pole.
b) The magnetic lines of force always emanate from the south pole and reach back the north pole moving inside the magnet.
c) Near the magnetic poles, the lines of force are closer. d) Near the magnetic poles, the lines of force are farther. 11. At any place, the lines of force in earth’s magnetic field are
a) Parallel b) Perpendicular c) equidistant d) at different distances 12. A positively charged particle (+q) of mass m enters a uniform magnetic field B with a velocity v perpendicular to the filed. The particle traverses a circular path. The radius of this circular path depends
a) linearly on the mass of the particle b) inversely on the mass of the particle c) linearly on the charge of the particle d) inversely on the charge of the particle
13. A positively charged particle (+q) of mass of m enters a uniform magnetic field B with a velocity v perpendicular to the field. The particle traverses a circular path. The radius of this circular path depends
a) linearly on the velocity of the particle b) inversely on the velocity of the particle c) linearly on the magnetic field d) inversely on the magnetic field
14. A proton and α and particle enter a uniform magnetic field with same speed perpendicular to the magnetic field. Choose the correct alternatives:
a) The radius of circular path of proton is larger b) The radius of circular path of α particle is larger c) The time period of proton is larger
d) The time period of α particle is larger.
15. The magnetic field generated at a point due to a small element of a current carrying conductor depends
a) linearly on the current flowing in the conduct. b) inversely on the current flowing in the conductor
c) linearly on the distance r between the element and point P.
d) inversely on the square of the distance r between the element and point P.
16. The magnetic field generated at a point due to a small element of a current carrying conductor depends
b) inversely on the length of the element
c) linearly on the sine on the angle between length of the element and the line joining the element to the point P.
d) inversely on the sine of the angle between the length of the element and the line joining the element to the point P
17. The magnetic force
a) does not have the same direction as magnetic field.
b) does not have the same direction as the direction of the motion of charges
c) direction is always perpendicular to both the direction of magnetic field and the direction of the motion of charges.
d) direction is always parallel to either the direction of magnetic field or the direction of the motion of charges.
18. A charged particle moves in a gravity-free spaced without change in velocity. Which of the following are possible ?
a) The electric field and the magnetic field both are zero. b) The electric field and the magnetic both are non-zero c) The electric field and is zero, the magnetic field is non-zero. d) The magnetic field is zero, the electric field is non –zero.
19. If a charged particle goes with a uniform velocity in a region containing electric magnetic field, then
a) electric field must be perpendicular to the magnetic field. b) electric field must be parallel to the magnetic field. c) the velocity must be perpendicular to the electric field. d) the velocity must be perpendicular to the magnetic field.
20. Let the two current carrying wires be placed near each other in vacuum or air. The force exerted on each other
a) depends linearly on the value of current in the wires. b) depends linearly on the length of the wires.
c) depends linearly on the distance between the wires. d) depends inversely on the distance between the wires.
21. The magnetic field at the centre of the current carrying circular coil is a) linearly proportional to the value of the current.
b) linearly proportional to the radius of the circular coil. c) inversely proportional to the radius of the circular coil.
d) inversely proportional to the number of turns in the circular coil. 22. Magnetic field inside a current carrying long solenoid
a) is independent of the radius of the solenoid. b) depends linearly on the radius of the solenoid. c) is independent of the number of turns of the solenoid. d) depends linearly on the number of turns of the solenoid.
23. When a charged particle moves in a uniform magnetic field at right angles to the direction of the direction of the field, which of the following quantities do not change ?
a) Path of the particle b) Speed of the particle c) Energy of the particle d) Charge of the particle. 24. Two freely hanging long wires are connected to a battery in
a) series , then the wires repel each other b) series, then the wires attract each other c) parallel, then the wires repel each other d) parallel, then the wires attract each other
25. A body is suspended from the lower end of a vertical spring. A current passes through the spring, then
a) the body shall be lifted upwards. b) the body shall go downwards.
c) the motion of the body depends on the direction of the current in the spring. d) the motion of the body is independent of the direction of the current in the spring.
26. A current is flowing in a circular loop of wire in clockwise direction. The magnetic field at the centre of the wire is
a) directed downward b) zero
c) inversely proportional to the radius of the loop d) directed upward.
27. An electron is moving in a liquid kept in a uniform magnetic field, in a plane perpendicular to the magnetic field. Then, the path of the electron is an inward spiral. The
a) kinetic energy of the electron decreases b) kinetic energy of the electron increases
c) frequency of revolution of the electron remains constant. d) frequency of revolution of the electron increases.
28. Two singly ionised isotopes of a material are accelerated through the same potential difference and enter perpendicular into a uniform magnetic field. Then,
a) the radius of the circular orbit of the heavier isotope in the magnetic field is greater than that of the lighter isotope.
b) the radius of the circular orbit of the heavier isotope in the magnetic field is less than that of the lighter isotope.
c) the kinetic energy of both the isotopes before entering the magnetic field are equal. d) the kinetic energy of both the isotopes will remain equal in the magnetic field. 29. An electron moving in a circular orbit around the nucleus of an atom
a) exerts an electric force on the nucleus
b) does not exert an electric force on the nucleus. c) has a magnetic dipole moment.
d) has zero magnetic dipole moment.
30. A bar of mass M, length L and carrying a current I is suspended horizontally by two wires in a uniform magnetic field B, which is directed into the page. Then,
a) the magnetic force ILB is directed downward. b) the magnetic force ILB is directed upward. c) the tension in each wire is (Mg+ ILB)/2. d) the tension in each wire is (Mg– ILB)/2
31. At a certain place, the component of the earth’s magnetic field parallel to the earth’s surface is 0.35 × 10–4 T. If an electron is shot with a speed of 107 m/x vertically upward at that place, then a) the force on the electron is 0.56 ×10–16 N
b) the acceleration on the electron is 6.1×1013 m/s2 c) the direction of force is eastward.
d) the direction of force is westward.
32. A proton is shot in the magnetic field 0.16i T with a velocity 105(i + 3j) m/x. Which of the following statements are correct ?
a) The path of the proton will be circular. b) The path of the proton will be helical. c) The radius of the path of the proton is 0.156 m. d) the pitch of the path of the proton is 0.164 m.
33. A particle with charge q and mass m orbits perpendicular to a uniform magnetic field. The frequency of orbital motion.
a) depends on the speed of the particle b) is independent of the speed of the particle c) depends linearly on q/m d) depends inversely on q/m
34. A uniformly charged disk of radius r, charge |q| and mass m with uniform mass distribution, rotates with uniform angular velocity w. Then,
a) the magnitude of magnetic dipole moment is w|q| r2/2 b) the magnitude of magnetic dipole moment is w|q| r2/4 c) the magnitude of angular momentum is m r2 w/2 d) the magnitude of angular momentum is m r2 w/4
35. A current I flows through a thin wire shaped as an equilateral triangle and as a regular hexagon, which are inscribed in a circle of radius R. The magnetic induction at the centre of the
a) equilateral triangle is 3 3µ0 I /2π R b) regular hexagon is 3 3µ0 I /2π R. c) equilateral triangle is 3µ0 I /π R d) regular hexagon is 3µ0 I / π R. 36. There are two rectangular wire frames A and B with same current I and same diagonal d. The
angle between the diagonals for frame A is 30°, while for frame B is 60°. The magnetic induction at the centre of frame
a) A is 8 µ0I / π d 3 b) B is 8µ0I /π d 3 c) A is 8µ0I / π d d) B is 8µ0I /π d.
KEY : (MULTI CORRECT ANSWERS)
1) a, d 2) a, d 3) a, b 4) a, b , c 5) b, d 6) a, c, d 7) b, c 8) a, b , d 9) a, d 10) b, c 11) a, c 12) a, d 13) a, d 14) b, d 15) a, d 16) a, c 17) a, b, c 18) a, b, c 19) a, c 20) a, b, d 21) a, c 22) a, d 23) b, c, d 24) a, d 25) a, d 26) a, c 27) a, c 28) a, c, d 29) a, c 30) b, d 31) a, b, c 32) b, d 33) b, c 34) b, c 35) a, d 36) b, c
MULTIPLE CHOICE QUESTIONS :
1. If an electron travels in a uniform magnetic field normally, the path of the electron is
1) straight line 2) ellipse 3) circular 4) parabola
2. If an electron traverses uniform magnetic field obliquely, its path is
1) a spiral 2) a circle 3) an ellipse 4) a parabola
3. A magnetic needle is kept in a non uniform magnetic field. It experiences
1) force 2) torque 3) torque and force 4) none
4. Check the correct statement
1) a spinning electron must have magnetic moment 2) every gas will be affected by magnetic field
3) electrically charged particles will be affected by magnetic field when they are at rest 4) neutron will be deflected in magnetic field
5. The force experienced by charge ‘e’ moving with velocity parallel to the magnetic induction B is
1) Be V 2) BV/e 3) Be/V 4) 0
6. Two particles of masses m1 and m2 (m1 > m2) have equal charges. When they enter the magnetic field at right angles to it with same K.E, the trajectory of first particle compared to that of second particle is
7. Positive particles coming from the space towards the earth are deflected by earth’s magnetic field towards
1) east 2) west 3) north 4) south
8. The force acting between two parallel current carrying conductors is F. If the current in each conductor and distance between them is doubled, the value of the force now is
1) F 2) 2F 3) 4F 4) 8F
9. The radius of curvature of the path of a charged particle in a uniform magnetic field is proportional to
1) charge of the particle 2) momentum of the particle 3) square root of its KE 4) both 2 and 3
10. If µo,εoand C are the permeability of free space, permitivity of free space and velocity of light, then
1) µoεo= 1/C 2) µoεo= 1/C2 3)
o
µ C =εo 4) µoεo= C
11. A coil of radius π meters, 100 turns carries a current of 3A. The magnetic induction at a point on its axis at a distance equal to 3 times its radius from its centre is
1) 7.2 x 10–6 wbm–2 2) 7.4 x 10–6 wbm–2 3) 7.5 x 10–6 wbm–2 4) 7.83 x 10–6 wbm–2 12. An α-particle of energy 2 MeV is moving in a circular path in a magnetic field. The energy
required by proton to move in the same circular path in the same magnetic field is
1) 0.5 MeV 2) 1 MeV 3) 2 MeV 4) 4 MeV
13. A vertical straight conductor carries a current vertically upwards. A point P lies in the east of it at a small distance and another point Q lies to the west of it at the same distance. The magnetic field at ‘P’ is
1) greater than at ‘Q’ 2) same as at ‘Q’
3) less than at ‘Q’ 4) depends on strength of current 14. An ammeter should have very low resistance
1) for large deflection 2) for better stability
3) so that it may not burn out 4) so that it may not change the value of current in the circuit 15. The resistance of an ideal voltmeter is
1) zero 2) 1000 Ω 3) infinity 4) 10 Ω
16. A voltmeter should have very high resistance
1) for large deflection 2) for better stability
3) so that it draws negligible current 4) so that it may not burnout 17. The moving of coil of galvanometer can be quickly stopped by
1) connecting a high resistance across the ends of the coil 2) holding a magnet near the coil
3) short circuiting the coil 4) earthing the coil
18. A volt meter has resistance of 1000 Ω. The additional resistance to be put in series to increase its range 5 times is
1) 5000 Ω 2) 200 Ω 3) 4000 Ω 4) 800 Ω
19. The sensitivity of a moving coil galvanometer depends on
1) the angle of deflection 2) earth’s magnetic field 3) torsional constant of the spring 4) moment of inertia of coil
20. Two coils one of 100 Ω and the other of 200 Ω are connected in series with a 4 V battery. A voltmeter of 200 Ω is connected in parallel with each of the coil is in turn. The voltage it shows in each case is
