29-7 The current in the long, straight wire AB shown in the figure is upward and is increasing steadily at a rate di/dt.
(a,b) At an instant when the current is i, what are the mag-nitude and direction of the field B at a distance r to the right of the wire? (c) What is the flux dΦB through the narrow shaded strip? (d) What is the total flux through the loop? (e) What is the induced emf in the loop? (f) Evalu-ate the numerical value of the induced emf if a = 12.0 cm, b = 36.0 cm, L = 24.0 cm, and di/dt = 9.60 A/s.
(a,b)
B = µ0i(t)
2πr into page (by right hand rule) (c)
dΦ = B dA =µ0i(t) 2πr L dr (d) Integrate
Φ = Z b
a
dΦ = µ0i(t)L 2π
Z b
a
dr
r = µ0i(t)L 2π ln b
a (e)
|E| = dΦ dt = µ0L
2π di dtlnb
a (f)
(2 × 10−7)(0.24 m)(9.6) µ
ln36 12
¶
= 5.06 × 10−7V What will be the direction of the current? Counterclockwise.
29-8 A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure. The field is changing with time, according to B(t) = (1.4 T)e−(0.057s−1)t. (a) Find the emf induced in the loop as a function of time (assume t is in seconds). (b) When is the induced emf equal to 1/20 of its initial value? (c) Find the direction of the current induced in the loop, as viewed from above the loop.
(a) We can find the emf induced in the loop using I
E· dl = −d dt
Z
B· dA = −dΦ dt,
where the integral of E · dl is the induced emf. Because the magnetic field makes a 60 degree angle with the loop, we can
write Z
B· dA = Φ = B(t)A sin(60)
where A = πr2 is the area of the loop. Because the area of the loop doesn’t change with time, dΦ/dt = πr2dB/dt, where
d
dtB(t) = (1.4)(−0.057)e−0.057t Putting these pieces together results in:
E = −πr2sin(60)(1.4)(−0.057)e−0.057t = 0.122e−0.057t We usually take E to be the magnitude of the emf, and don’t worry about the sign. The negative sign is related to the polarity of the voltage.
(b) Now that we have an expression for the induced emf as a function of time, we can solve for t at a particular emf.
1
20 = e−0.057t =⇒ t = ln 20/(0.057 s−1) = 52.6 s (c) According to Lenz’s Law, the induced emf will cause current to flow in the direction that opposes the change in magnetic flux. Because the magnitude of the magnetic field directed upwards through the loop is decreasing with time, the induced emf will cause a current to flow in a direction that will cause a magnetic field upward. For that to happen, the current must flow in a counterclockwise direction.
October 27, 2011
29-11 In a region of space, a magnetic field points in the +x direction (toward the right). Its magnitude varies with position according to the formula Bx = B0+ bx, where B0
and b are positive constants, for x ≥ 0. A flat coil of area A moves with uniform speed v from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin?
(b) As viewed from the origin (which we take to be to the left), what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answer to part (a)? (d) If instead the coil moved from left to right, what would be the answer to part (b)?
+X B(x) = B + bx
0v
(a) This problem should be solved with Faraday’s Law:
E = −d dtΦ.
The area of the loop remains constant, but the magnetic flux will change with time because the loop moves towards a weaker field.
Φ = A (B0+ bx)
Using the following equation, we can get the time derivative of the flux at the location x of the loop:
dΦ dt =dΦ
dx dx
dt = (Ab)(−v) = −Abv.
Now the emf can be calculated using E = −dΦ
dt = Abv.
(b) The direction of the current can be determined with Lenz’s Law. The direction of the induced current will be in a direction to oppose the change in flux. Because the flux through the loop is decreasing, the induced current will in-duce a magnetic field inside the loop pointing to the right.
This means, as viewed from the origin (on the left), the cur-rent will be going in a clockwise direction (CW).
(c) If the coil is moving from left to right instead, the problem is similar to part (a), except the velocity is positive, and so the answer is −Abv. Since no sign convention is defined for the emf enter Abv for Mastering Physics.
(d) Because the coil is moving in the opposite direction com-pared to parts (a) and (b), the flux through the loop is in-creasing, and so the induced current will induce a magnetic field pointing towards from the origin. So, viewed from the origin (on the left), the current will be in a counter-clockwise direction (CCW), the opposite direction from part (b).
29-13 The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of 1.95 cm.
The coil rotates in a magnetic field of 7.80× 102T. (a) What is the angular speed of the coil if the maximum emf produced is 2.60 × 102V?
(a) The figure below shows the general configuration of the square coil and the magnetic field. Note that our coil has 120 turns. The magnetic field is constant and to the right, while the coil (and consequently, the vector A perpendicular to the loop) changes with time.
The angle φ between the magnetic field B and the vector A is given by φ = ωt. Thus dφ/dt = ω, and the flux through the coil at any given time can be written as,
ΦB= B · A = BA cos φ = BA cos ωt.
¿From Faraday’s law, the induced emf is given by the change of the flux with respect to time. But here, we must remember we have a coil with N = 120 turns, thus,
E = −NdΦB
dt = ωN BA sin ωt
The maximum induced emf occurs when sin ωt = 1. We can solve for the angular speed that would create the stated emf:
Emax= ωN BA ω = Emax
N BA = 2.60 × 102V
120(7.80 × 102T)(0.0195 m)2 = 7.31rad s
29-17 Consider the system shown below. (a) Using Lenz’s law, determine the direction of the current in resistor ab of the figure when switch S is opened after having been closed for several minutes. (b) Using Lenz’s law, determine the direction of the current in resistor ab of the figure when coil B is brought closer to coil A with the switch closed. (c) Using Lenz’s law, determine the direction of the current in resistor ab of the figure when the resistance of R is decreased while the switch remains closed.
(a) The current will flow from the positive terminal of the battery when S is closed. Thus, the current flowing in coil A will produce a magnetic field inside coil A pointing to the right (and also to the right inside coil B). When S is then opened, the current will stop flowing and the magnetic field it produces will decrease. Lenz’s Law says that the E induced in the coil B will oppose the change in magnetic flux by causing current that will create an additional magnetic field to the right. Thus, by the right hand rule the current in coil B will flow from a to b.
(b) When S is closed, coil A produces a magnetic field inside the coils pointing to the right. As coil B is brought closer to A, the magnetic field inside coil B becomes stronger, so the flux becomes larger. The E induced in coil B will oppose the change in magnetic flux by causing current that will create additional magnetic field to the left. By the right hand rule, the current in coil B will flow from b to a.
(c) If the resistance in R is decreased, the relation I = V /R tells us that the current will increase, since the battery volt-age V does not change. When the current through coil A increases, the magnetic field directed to the right will also increase. The E induced in coil B will oppose the change in magnetic flux by causing current that will create additional magnetic field to the left. By the right hand rule, the current in coil B will flow from b to a. [This argument is exactly the same as the one used in part (b)].
29-20 A 1.60 m long metal bar is pulled to the right at a steady 4.5 m/s perpendicular to a uniform, 0.755 T magnetic field. The bar rides on parallel metal rails connected through R = 26.0 Ω, as shown in the figure, so the apparatus makes a complete circuit. You can ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit. (c) Calculate the current through the resistor.
The flux at any time is the area of the loop times the mag-nitude of the magnetic field B, and we let L be the length of the bar. In a time ∆t the area of the circuit will increase by L∆x, where ∆x and ∆t are related to the speed of the bar by v = ∆x/∆t.
(a) The emf is given by
E = ∆Φ/∆t = BL∆x/∆t = BLv
= 0.755 T × 1.60 m × 4.5 m/s = 5.44 V
(b) Pulling the rod increases the flux by adding magnetic field directed into the page. By Lenz’ Law, the induced current must then produce a magnetic field inside the loop directed out of the page. The current must therefore be counterclockwise.
(c) By Ohm’s Law,
E = IR ⇒ I = 5.44 V/26.0 Ω = 0.21 A