Chapter 30: Faraday’s Law
7. We take t = 0 when the leading edge of the loop enters
the magnetic field. The position of this edge is given
by x = – L + vt. The leading edge will reach the z-axis
at t1 = L/v and the other side at t2 = 2L/v. The trailing
edge will leave the magnetic field at t3 = 3L/v.
We have three regions to consider:
Region I, 0 < t < (L/v).
The magnetic flux through the loop is
ΦB = BL(x + L). (Note that x < 0.)
The induced emf is
ε1
y
r
B
L
r
v
x
L
L
r
B
L
z
= – dΦB/dt
= – BL(dx/dt) = – BLv (clockwise as viewed from above).
Region II, (L/v) < t < (2L/v).
The magnetic flux through the loop is
ΦB = [+ B(L – x) – B(x)]L = BL(L – 2x). (Note that x > 0.)
The induced emf is
ε2
= – dΦB/dt
= – BL(– 2 dx/dt) = + 2BLv (counterclockwise).
Region III, (2L/v) < t < (3L/v).
The magnetic flux through the loop is
ΦB = – B(2L – x)L = B(x – 2L)L. (Note that x > L.)
The induced emf is
ε3
= – dΦB/dt
= – BL(dx/dt) = – BLv (clockwise).
10. If we take ther area
r of the coil to be aligned with the field at t = 0, we have
ΦB = N B · A = NBA cos(ωt) = NBA cos(2πft).
The emf induced in the coil is
ε = – dΦB/dt = – N (d/dt)[BA cos(2πft)]
= NBA2πf sin(2πft).
The current in the coil is
I = ε/R = NBA2πf sin(2πft)/R.
The maximum value occurs when sin(2πft) is maximum:
Imax = NBA2πf/R
3.0 A = (450 turns)(0.35 T)π(2.5 × 10–2 m)22πf/(12 Ω), which gives f = 19 Hz (1.6 × 102 rad/s).
12. At t = 0, the flux through the loop is maximum. The rate of change
r
ω
of the flux, however, is zero, so ε = 0.
At t = T/4, the flux through the loop is zero. The flux is changing
from entering one side of the loop to entering the other side. The
induced current will oppose this change by producing a magnetic
field that will maintain the original flux. For the view in the
figure, the induced field is directed into the page, and ε is clockwise.
At t = T/2, the flux through the loop is maximum. The rate of change
of the flux, however, is zero, so
ε = 0.
At t = 3T/4, we have the same view as for t = T/4.
ε is clockwise.
1
Rotation
r
B
x
Chapter 30: Faraday’s Law
Note that, in the reference frame of the loop, the emf has reversed.
r
B0
16. From the expression for the radius, r = r0(1 + αt), we have
dr/dt = r0α.
Because the magnetic field is constant and perpendicular
to the ring,rwe rfind the flux through the loop from
ΦB = B · A = B0A = B0πr2.
The induced emf is
ε = – dΦB/dt = – 2B0πr (dr/dt) = – 2B0π[r0(1 + αt)]r0α
r
clockwise.
The resistance of the ring is
R′ = R2πr = R0(1 + βt)2πr0(1 + αt),
so the induced current is
I = ε/R′ = – 2B0πr02α(1 + αt)/[R0(1 + βt)2πr0(1 + αt)]
= – B0r0α/R0(1 + βt) clockwise.
24. We find the flux through the area swept out by a radial line on the disk, as in Example 30–7:
ΦB = B(1/2 θR2) = ½ BωtR2.
The induced voltage is
ε = – dΦB/dt = – ½
BωR2
= – ½ (0.10 T)(150 rad/s)(0.035 m)2 = – 9.2 × 10–3 V.
r
r
The negative sign indicates that when the direction of ω is the same as the direction of B , the emf is
directed radially out from the axis, and the rim is at the higher potential.
27. (a) FB = evB = (1.6 × 10–19 C)(0.25 m/s)(0.030 T) = 1.2 × 10–21 N .
(b) Let FE = eE = eV/d = FB to obtain
V = FB d/e = evBd/e = vBd = (0.25 m/s)(0.030 T)(0.003 m) = 2 × 10–5 V.
2
Chapter 30: Faraday’s Law
34. The power supplied by the force equals the power dissipated
by the current in the resistance of the loop. We take t = 0
when the front tip of the loop is leaving the region of the
magnetic field. The position of the front tip is x = vt.
For 0 < t < L/v√2, the first half of the loop is leaving the field.
The magnetic flux through the loop is
ΦB = BA = B(L2 – x2).
The induced emf is
r
B
ε = – dΦB/dt = B2x (dx/dt) = 2Bxv = 2Bv2t.
The power dissipated in the resistance, which must be the power
supplied by the external force, is
P = I 2R = ε 2/R = (2Bv2t)2/R for 0 < t < L/v√2.
For L/v√2 < t < (2L/v√2 = L√2/v), the second half of the loop is leaving the field.
The magnetic flux through the loop is
ΦB = BA = B[(2L/√2) – x]2 = B(L√2 – x)2.
The induced emf is
ε = – dΦB/dt = – B2(L√2 – x) (– dx/dt) = 2B(L√2 – x)v = 2Bv(L√2 – vt).
The power dissipated in the resistance, which must be the power supplied by the external force, is
P = I 2R = ε 2/R = [2Bv(L√2 – vt)]2/R
35. (a)
for L/v√2 < t < L√2/v.
If L is the separation of the rails and D is the distance of the bar from the end of the rails, the
magnetic flux through the loop is
ΦB = BA = BLD.
The induced emf is
ε = – dΦB/dt = – BL (dD/dt) = – BLv,
which produces a current
I = ε/R = BLv/R.
The Lorentz force on this current is
F = ILB = B2L2v/R, opposite to the direction of the motion.
To maintain the motion, an equal and opposite external force is required:
F = (0.28 T)2(0.30 m)2(0.60 m/s)/(0.050 Ω) = 0.085 N.
(b) The rate of Joule heating in the resistor is
P = I 2R = (BLv/R)2R = B2L2v2/R = Fv
= (0.085 N)(0.60 m/s) = 0.051 W.
3
L
L
x = vt
Chapter 30: Faraday’s Law
50. From the cylindrical symmetry, we know that the electric
field depends only on r and must be circular.
For path 1, r < R, the magnetic field through the path is
B = μ0nI, so the flux is
ΦB = BA = Bπr2 = μ0nIπr2.
If we apply
Faraday’s law, we have
r
r
∫ E · d s = – dΦB/dt; E ∫ ds = – μ0nπr2 (dI/dt);
E2πr = – μ0nπr2I0ω cos(ωt), which gives
E = – (μ0nI0ωr/2) cos(ωt).
For path 1, r = R/2; we have
E = – (μ0nI0ωR/4) cos(ωt) circular.
For path 2, r > R, the magnetic field through the path
is B = μ0nI inside the solenoid only, so the flux is
ΦB = BA = BπR2 = μ0nIπR2.
If we apply
Faraday’s law, we have
r
r
∫ E · d s = – dΦB/dt; E ∫ ds = – μ0nπR2 (dI/dt);
I
E2πr = – μ0nπR2I0ω cos(ωt), which gives E = – (μ0nI0ωR2/2r) cos(ωt).
For path 2, r = 3R; we have E = – (μ0nI0ωR/6) cos(ωt) circular.
4
Path 1
Path 2