Phy 1 (15)

Q.1. 2 1 R I( k)ˆ     U1 = -1.B0   2 2 R I(i )ˆ    U2 = - 1.B R IB2 0   U = R2IB0 = 2 2 1 mR 2 2         = 2 IB0 m  Q.2. Fm  B  a  B a.B  0 ˆ ˆ ˆ ˆ ( 8 / 3i yj).(3i4 j)0 y = 2 Q.3. B = R 2 I R I 2 . 4 0 0  _     = 0I 2R  2  Mutual inductance M = R 2 I 2 0    .

Q.4. If T is the tension in the loop, for equilibrium of a small part of it

2T sin  = dF = BidL

for an element sin    and d2 = 2R,

2T = BI  2R  T = BIR [L = 2R] T =  2 BIL = 0.125N 14 . 3 2 5 . 0 57 . 1 1     Current is clockwise.     dF T R T

Q.5. If the particle is projected perpendicular to the Bfield. The angular velocity,  is

qB m     B 3 17 kˆ 3 jˆ 2 iˆ 2 17 _        _{ } 

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2 i 2 j 3k B 3         tesla Q.6. Fm  B  a  B a.B 0 ˆ ˆ ˆ ˆ ( 8 / 3i yj).(3i4 j)0 y = 2 Q.7. B = oni n B i turns m o           1 4 10 1 1 4 10 7 7 / B = oni At the ends B = 2 ni   Q.8. I = I0 (1 – e-Rt/L) …(i) U = LI2 2 1 , Umax = LI2₀ 2 1 U = Umax 4 1 2 0 2 _LI 8 1 LI 2 1   I = I0 / 2 From (i) 2 I₀ = I0 (1 – e -Rt/L )  t = ln2 R L = 5 ln2 = 3.47 s

Q.9. _{Iabj Ibc cos 45ˆ} 0

 

_ˆj

    Ibc sin 45 i0ˆ 1 _{ˆ ˆ} 1 _ˆ 1 4 2 i 1 12 j 8 j 2 2       





2 ˆ ˆ 4 2 i 4 3 2 j A m    Magnitude 2 16 2 16 6.7 11.8 Am       Q.10. 0 1 2 3 I 1 _ˆ 1 _ˆ 1 _ˆ B k i j 8 R R R     _   _    0 2 2 2 1 2 3 I 1 1 1 B 8 R R R     Q.11. Fm  B  a  B a.B  0 ˆ ˆ ˆ ˆ ( 8 / 3i yj).(3i4 j)0 y = 2

Q.12. || = MBsin  = niAB sin   = (0.04) (2) sin 900 = 0.25 N- m Q.13. Fm  B  a  B a.B  0 ˆ ˆ ˆ ˆ ( 8 / 3i yj).(3i4 j)0 y = 2

Q.14. Magnetic induction at point A,

B1 = _{2 2 3/2} 2 0 ] d r 2 [ ) r 2 ( I 2 4     and due to coil B,

B2 = _{2 23/2} 2 0 ] ) 2 / d ( r [ Ir 2 . 4     [ r = dtan

2  where 2 = solid angle]

 2 1 B B 2 1  .

Q.15. (a) The loop must wind around the 1A wire twice as many

times as it winds around the 2A wire, but in the opposite sense.

1A 2A

(b) In this case the net current linked by the loop is found to be

4A: 7 7

16 10 4 10 4

     

One way to get this would be to go around the 1A wire twice and the 2A wire once in the same sense

1A 2A Q.16. B = 0 0 1 2 i i 4 r 4 r        0 1 2 1 2 i r r 4 r r      _{ }   . Q.17. | | = MBsin  = niAB sin   = (0.04) (2) sin 900 = 0.25 N- m

Q.18. The given loop can be considered as combination of the two loops as shown in figure.

(2) (1) 2 z x y 10 cm 40 cm 25 cm

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jˆ iA kˆ iA P P P  ₁₂  ₁  ₂ Where A1 = 0.25  0.4 = 0.1 m 2 A2 = 0.40  0.10 = 0.04 m2



0.2kˆ 0.08jˆ

 

0.2iˆ 0.5jˆ 3kˆ



B P           = -



0.34iˆ0.04jˆ0.016kˆ



Nm Q.19. (a) mv = 2Em = 8  910  10-25  = m qB  T = qB m 2

For minimum value of B electron will strike S after one full rotation GS = V|| T  0.1 = v cos  qB m 2  B = e 1 . 0 m 2 cos V     = 4.74  10-3 T. Q.20. Let BB₁iˆB₂jˆB₃kˆ 

Hence q



viˆ



B₁iˆB₂jˆB₃kˆ





qvB



3jˆ4kˆ



 B₂kˆB₃jˆ3Bjˆ4Bkˆ  B2 = 4B, B3 = 3B

Also q



vjˆ



B₁iˆB₂jˆB₃kˆ





qvB

 

3iˆ  B₂kˆB₃iˆ3Biˆ  B1 = 0 Hence B 4Bjˆ3Bkˆ  kˆ 5 3 jˆ 5 4 B B B     

Q.21. (a) At P due to current in (1), magnetic field is in upward direction and due to current in (2),

magnetic field is downward direction.

At Q due to current in (1) magnetic field is downward and due to current in (2), magnetic field is also downward direction.

Therefore at P, B1 - B2 = 3 . 0 30 2 1 . 0 20 2 o o     

= 2  10-5 N/Ampmeter, along positive z-axis at Q, B1 + B2 = 3 . 0 30 2 1 . 0 20 2 o o     

= 1  10-4 N/Ampmeter, along negative z-axis at R, B2 - B1 = 3 . 0 20 2 1 . 0 30 2 o o     

= 4.7  10-5 N/Ampmeter, along positive z-axis

Q.22. d dx.v. 0I 2 x     a L o a Iv dx 2 x    



0Iv _n a L 2 a      _{ }    

Q.23. B = B1 + B2 = 2 a 2 I . 4 2 3 a I . 4 0 0        = 0 10 5 3 7 2 1 3 a 8 I             T

Q.24. F IL B = ˆj× [i+ j+ k]ˆ ˆ ˆ Newtons = (ˆik)ˆ Newtons.

Q.25. (a) The loop must wind around the 1A wire twice as many

times as it winds around the 2A wire, but in the opposite

sense. 1A 2A

(b) In this case the net current linked by the loop is found to be

4A: 7 7

16 10 4 10 4

     

One way to get this would be to go around the 1A wire twice and the 2A wire once in the same sense

1A 2A

Q.26. The force is given by the vector relation

B      i F

 F = iB sin where  is the angle between  andB

F = (3.0A)  (5  10-2 m)  (10-3 Wbm-2)  0.5 = 7.5  10-5 N

The direction of this force is perpendicular to the plane which contains both  and B.

Q.27. F I(LB)

The angle between F and B is 1800 The angle between BA and B

 is 00 For both these LB0

 

For ED, FED  I(LB)  ILBsin900 iˆ  ILBiˆ    Similarly, FCB I(L B)    

 = ILB sin 2700 = - ILB iˆ  FED FCB 0   ) B L ( I

F_DC   = I[LiˆBjˆ] = ILB kˆ  FABFBCFCDFDEFEF ILBkˆ     

 The magnitude of the two required force is ILB and it is directed along +ve z-axis.

Q.28. The force is given by the vector relation

B      i F

 F = iB sin where  is the angle between  andB

F = (3.0A)  (5  10-2 m)  (10-3 Wbm-2)  0.5 = 7.5  10-5 N

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Q.29. The repulsive force on the side ps of the current-carrying loop, due

to current i1 is N 10 4 . 2 04 . 0 15 . 0 16 20 ) 10 2 ( d L i i 2 F o 12 7 4 1            

This force will be towards RHS and  to the current-carrying wire ps.

Similarly, the attractive force acting on the side qr of

the loop, due to current I1 is(Here R = d+b = 10 cm = 0.1 meter)

10 . 0 15 . 0 16 20 ) 10 2 ( F₂   7      N 10 96 . 0  4  . a q p s i2 i1 b r d

Direction of this force will be towards LHS and  to current-carrying wire qr. The forces acting on the sides pq and rs of the loop will be equal and opposite. Thus net force on the loop = F1  F2 = (2.4  0.96)  104 = 1.44  104 N

(Acting away from the current-carrying wire)

When the direction of current in the loop becomes clockwise, the net force on the loop remain same, but its direction now becomes towards the current-carrying wire.

Q.30. Here the angle between v and B is  = 450 hence the path is a helix

the axis of the helix is along x-axis (parallel to B).

Q.31. (a) Field due to a single coil (along x)

= ( i) R 5 5 I 4 ) i ( ) 4 / R R ( 2 IR 0 2 / 3 2 2 2 0        

Field due to infinite wire 0I

2 R  

 j



Total magnetic field = ( i) R 5 5 I 4 0    + 0I 2 R   j 

Q.32. A conducting rod 'OA' of mass 'm' and length 'l' is kept rotating in a vertical plane . . . any other

resistance. (a) 2 1 Bl2 = e (b) E = iR + L dt di 

_

_

  iR E di L dt   log(E iR) c L Rt     E  iR = EeRT/L  i =



1 e RT/L



R E _  i = Bl2



1 e RT/L



2 1 R 1 _         i = R 2 Bl2 at t   steady state

Power = Torque () i2 R = J J =     2 2 4 2 2 R 4 R l B R i

+ torque due to weight of the rod

J = R 4

l B24

+ torque due to weight of the rod = R 4 l B24 + Mg (/2)cos t Q.33. (a) B = B1 – B2 = ( ) r 2 I . 4 ) ( r I . 4 0 0        = r 8 I 2 1 1 r 4 I ₀ 0           (b) B = B1 + B2 = 2 a 2 I . 4 2 3 a I . 4 0 0        = 0 10 5 3 7 2 1 3 a 8 I             T

Q.34. If I1 & I2 be the currents in circular and straight part

respectively & B1, B2 the magnetic fields due to them,

then B1 = R 3 I 3 2 R 2 I_{1 01} 0 _{ }   B2 = R 2 I 3 ] 60 sin 2 [ ] 60 cos R [ 4 I 0 _{0 2} o 2 0      R I2 I I I1 240o O

For the total field at 'O' to be zero R 2 I 3 R 3 I_{1 0 2} 0        2 3 3 I I 2 1 Now, 2 2 1 1 2 2 1 1 2 2 2 1 2 2 1 r r r r R R I I                    Required ratio = 2 1 2 1 I I            2 = 2R sin 60o = 3R, 1 = R 3 4 3 4 R          3 3 4 2 1      Required Ratio = _                  3 3 4 2 3 3 = 2 .

Q.35. If r be the resistance per unit length the effective resistance ‘R’ of the half loop, connected across A

and B will be R = _                       ) a b ( 2 a 3 2 b 3 r R = 5.6 .

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Hence, I (current in the loop as indicated) =

R V_ba

= 5A. If  be the torque on one- half of the loop

connected across A & B, total torque = 2  = 2

_

          b a 0 r 2 I )} 6 / sin( r 2 { I dr (- iˆ ) = - 0II'   (b – a) iˆ = -7.5  107 iˆ N-m. O /6 A B 2r sin /6 dr r Q.36. F q(vB) For the first case

–(52  103 N) kˆ = (105 C) m/s (iˆ jˆ) (B iˆ B jˆ B kˆ) 2 10 z y x 6             = [B iˆ B jˆ (B B )kˆ] 2 10 x y z z           Bz = 0, By - Bx = -103 T . . . . (1)

For the second case

 jˆ F_y (10-5C) (106 m/s) (kˆ )  [(BxiˆB_yjˆB_zkˆ)]  jˆ F_y 10(B_xjˆB_yiˆ)  Fy = 10 Bx , By = 0  using (1) we get Bx = 103 T Thus B = (103 T) iˆ Also Fy =10 Bx = 102 N.

Q.37. The surface charge density is q/r2. Hence the charge within a ring of radius R and width dR is

) RdR ( r q 2 ) RdR 2 ( r q dq ₂   ₂  

The current carried by this ring is its charge divided by the rotation period,

] dR . R [ r q / 2 dq di ₂      

The magnetic moment contributed by this ring has the magnitude dM = a |di|, where a is the area of the ring.

Q.38. Separation = 2 sin /2 = x (say)  Force = ) 2 / sin 2 ( 2 i x 2 i 02 2 0          … (i) downward force = mg = ()g … (ii)  tan /2 = g ) 2 / sin 4 ( i F F ₀2 G E          /2 /2 F mg Q.39. (a)            2 2 3 R I 4 | B

| 0 _{in outward direction of plan of paper.}

Q.40. F q(v)(B_xiˆB_yjˆB_zkˆ) Case- I when v is along x. we get By = B 2 3 , Bz = +B/2 Case – II We get, Bx = 0, Bz = B/2  kˆ 2 1 jˆ 2 3 Bˆ  

Q.41. (a) Magnetic moment due to loop KLM = ( ) 2 2 i R 



Magnetic moment due to loop KNM = (j) 2 R2 

net magnetic moment m = [ i j] 2 R2    (b) = mB = ( i j) (3i 6j 3k) 2 R2       = [ 3i 3j 9k] 2 R2    

Torque about x-axis is [ 3j 9k] 2 R2    .

Q.42. Let A = acceleration & v = instantaneous velocity.

3mg – T = 3m A & T- Fm = mA where Fm = vl 2

B2/ R = V ( putting the values) Integrating , v = 3g [ 1 – e- t/4 ] Q.43. [ i ]Force on AD = 0I1 I2/ 2 Force on AB = ln2 2 I I x / dx I I 2 2 1 0 a 2 a 2 1 0     



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[ii ] Net torque about BC = 0 as the force passes through the axis of rotation & torque due to forces on AB & CD cancel one another.

Angular acceleration = 0  The required time is infinite.

Q.44. For A, ) kˆ ( d 2 I 2 ) A at P to due ( B 0 1      ) jˆ ( d 2 I ) A at Q to due ( B 0 2     ) jˆ ( ) d 3 ( 2 I ) A at R to due ( B 0 3      Resultant field B B₁ B₂ B₃        = _                       ) kˆ ( 2 jˆ 3 2 d 2 I jˆ ) 3 1 ( jˆ ) kˆ ( 2 d 2 I ₀ 0 For B, ) kˆ ( d 2 I 2 ) B at P to due ( B 0 1     ) jˆ ( ) d 3 ( 2 I ) B at Q to due ( B2 0     ) jˆ ( d 2 I ) B at R to due ( B 0 3      Resultant field B B₁B₂B₃ = _                     kˆ 2 ) jˆ ( 3 2 d 2 I jˆ jˆ 3 1 kˆ 2 d 2 I ₀ 0 Q.45. Fq

 

v 



B_xiˆB_yjˆB_zkˆ



Case - I When V V_i    We get BY 3B, Bz 4B Case - II When V V_j we get Bx 0 Bz 4B  kˆ 5 4 jˆ 5 3 | B | B Bˆ     

Q.46. (a) B1 field due to a single coil (along x) =

R 5 5 Ni 4 ) 4 / R R ( 2 iNR ₀ 2 / 3 2 2 2 0 _   

Field due to both coils , B0

R 5 5 Ni 8₀  A B R x B2 + B1 = 2B1 = 4.49  10 -3 T

(b) If a particle has the velocity v = 106          jˆ 2 3 iˆ 2 1

and the field is B = iˆ R 5 5

Ni

80 _{.& the force, kˆ}

2 3 B qv B v q F _ _ _ 0 0 _, = 3.89  10-3 N. Q.47.  =

_

             x a x 0 0 x x a ln 2 Ia ady y 2 I =          vt a 1 ln 2 Ia 0  =





2 0 B vt a vt a vt 2 Ia dt        I =



a vt



t 8 Ia 0    dy y x _v I Q.48. L Bv dt di  L dt dx B dt di   i = x L B

Magnetic force on the rod, Fm = iB = x L B22



m v C B A D  m x L B dt x d 2 2 2 2 _ 

 (Force is in opposite direction of v)

2 2 2 2 2 d x B dt  m L  x  = mL B .

Q.49. Let the current be in the outer coil.

The field at centre B = 0I

2b 

The flux through the inner coil =

2 0I a

2b  

The induced emf produced in the outer coil  = d dt  





2 2 2 0 a d 2 0 a t 2t 2b dt b     

Current induced in the outer coil =

2 0 2 a t R bR    

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Heat developed in the outer coil =

t 2 0 I Rdt  = 2 2 4 2 t 0 2 2 0 4 a t Rdt b R    = 2 2 4 3 0 2 4 a t b R 3  

Q.50. (i) Velocity of wire frame when it starts entering into the magnetic field.

V1 = 2gh = 2

  

10 5 = 10m/s.

And the time taken is t1 =

 

1s 10 5 2 g h 2  

(ii) When the frame has partially entered into the field, the induced e.m.f. produced is  = Blv I =  R B v R  l (anti - clockwise) Ampere’s force F = B v R 2 l2 (upward)                                     mg F l

Net acceleration downward = g - mR v B2 2  Putting m = 0.5 kg, B = 1 T, l = 0.25 m, v = 10 m/s, R = 1/8  We get, F =

  

  

5N 8 / 1 10 25 . 0 12 2  Since, mg = (0.5)(10) = 5 N

Therefore, using Newton’s second law , the acceleration of the wire frame while entering into the magnetic field is zero. Thus time taken to completely enter into the field is

t2 = 0.2s

10 2



(iii) When the frame has completely entered into the field, the current becomes zero and thus, the ampere’s force also become zero. The frame accelerates under gravity only.

 15 = 10t3 + 5 t3 2

or t₃2 2t₃   3 0 or t3 = 1s

The total time taken is

T = t1 + t2 + t3 = 1 + 0.2 + 1 = 2.2 s Q.51. Kinetic energy of ions =

2 1

Mv2 = eV . . . (i)

And also Bev =

R Mv2

. . . (ii) combining (i) and (ii)

The radius of the path traversed : R =

e B VM 2 e B VeM 2 2 2 2 

In mass spectroscope U236 ion will follow a semicircular path of radius R1 = e B VM 2 2 236 and radius of U239 , R2 = e B VM 2 2 239

Separation between the pouches S = 2(R2 – R1) S = 2          e B VM 2 e B VM 2 2 236 2 239 = 1.98  10-2 _Wb/m2_. U236 U239 S Q.52. (a) 1 1 1 0 1 nˆ r 2 I B     , 2 2 0 0 2 nˆ r 2 I B     , Since I2 is absent  B B1 B2     

As B₁ and B₂ are oppositely directed nˆ ] B B [ B 1 2  = r nˆ Ar 1 r Ar 1 2 2 2 2 2 1 1 0            = nˆ A 2 Ia 0 

where, A = R2, and a is the separation of the two centres.

(b) For any point p inside the cavity, if r₁ and r₂ be the position vectors of p with respect to C1 and

C2 then, magnetic field at p would be

] r J [ 2 ] r J [ 2 2 0 1 0 _ _ _  = [r r ] 2 1 2 0 _ 

[where J is the current density vector] = 0 _nˆ1

A Ia 2 

Q.53. The electric force

  Eq F_e  1.6  10-19  102.4  103

 

kˆ



v B



q F_B   = 1.6  10-19  1.28  106  8  10-2

 

kˆ = 1.6  10-19  102.4  103

 

_kˆ so net force FFeF_B = 0  a = 0

Hence the particle will move along + x axis with constant velocity v = 1.28  106 _{m/s so the}

distance traveled by the particle in time 5  10-6 sec. x0 = vt = (1.28  106)  (5  10-6) = 6.40 m

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Q.54. Charge contained by a ring of radius x and

thickness dx is dq = 2x dx 

The equivalent current i = T dq =   / 2 dq and the corresponding magnetic moment d = iA =   2 dq x2 =  x3 dx  x

 Total magnetic moment  =  d = 4 1  R4 similarly dB = x 2 i 0  =    2 x 2 0 _{2x dx }  B = dB = R 2 1 0        .

Q.55. (a) As the initial velocity of the particle is perpendicular

to the field the particle will move along the arc of a circle as shown.

If r is the radius of the circle, then

0 0 2 0 _{qv B} r v m 

Also from geometry, L = r sin 30°  r = 2L or L = 0 0 B q 2 v m 300 300 v0i r r B0k x = L (b) In this case L = r B q 2 mv 1 . 2 0 0 _

Hence the particle will complete semi-circular path and emerge from the field with velocity –v0 iˆ as shown.

Time spent by the particle in the magnetic field

0 v r T  = 0 B q m 

The speed of the particle does not change due to magnetic field.

L >r

-v0i

Q.56. For conductor with current into the plane

for a circle of radius a in the cross-section of conductor current flowing within it

I =

_

 a 0 xdx 2 0 (x/R) = 3 a R 2₀ 3 O P B B0



d-a a

For all point on the circle of radius a due to symmetry B is same. Applying Ampere's law



a 2 0 d . B  = 0I  B0 = 3 a R 2 a 2 3 0 0   

At P, field B is  to OP upward as shown

Now field B0 due to wire must balance it, B0 must be downward opposite to be B of same

magnitude  current in wire should also be into the plane of paper, Let it be I0 .

B = a d I 2 4 0 0    0 B B = 0 | B | | B | ₀    3 a R 2 2 a d I 2 4 2 0 0 0 0         I0 = a (d a) R 3 20 2 _{ .}

Q.57. (a) Area vector of loop A =

           2 kˆ 2 iˆ ) 1 ( 2 2

Area vector of loop B =

           2 kˆ 2 iˆ ) 1 ( 2 2

Hence net vector dipole moment = 10  5  /2            2 kˆ 2 iˆ + 10  5  /2            2 kˆ 2 iˆ = - iˆ 2 25 A-m2

(b) Net torque acting on the system : B M      = iˆ



iˆ jˆ kˆ



2 25           = (jˆ kˆ) 2 25   Nm. B = a d I 2 4 0 0    0 B B = 0 | B | | B | ₀    3 a R 2 2 a d I 2 4 2 0 0 0 0         I0 = a (d a) R 3 20 2 _{ .}

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Q.58. Charge particle will move from outside of the ring when

x-coordinate will be maximum and x  R0

i.e. x = R[1 – cos ] = R[1 – cos ] = 2R xmax = 2R = 2 mv sin qB  i.e. t =  i.e. t = m/qB z = v cos   t = v cos  m qB  Hence z  v cos   R 2v sin     2R cot0   R  x x y