Problem#1
A small, circular ring is inside a larger loop that is connected to a battery and a switch, as shown in Fig. 1. Use Lenz’s law to find the direction of the current induced in the small ring (a) just after switch S is closed; (b) after S has been closed a long time; (c) just after S has been reopened after being closed a long time.
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| Fig.1 |
Answer:
Lenz’s law requires that the flux of the induced current opposes the change in flux.
(a) ΦB is ⨀ and increasing so the flux Φind of the induced current is ⨂ and the induced current is clockwise.
(b) The current reaches a constant value so ΦB is constant. dΦB/dt = 0 and there is no induced current.
(c) ΦB is ⨀ and decreasing, so Φind is ⨀ and current is counterclockwise.
Only a change in flux produces an induced current. The induced current is in one direction when the current in the outer ring is increasing and is in the opposite direction when that current is decreasing.
Problem#2
A circular loop of wire with radius r = 0.0480 m and resistance R = 0.160 Ω is in a region of spatially uniform magnetic field, as shown in Fig. 2. The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 8.00 T and is decreasing at a rate of dB/dt = –0.680 T/s. (a) Is the induced current in the loop clockwise or counterclockwise? (b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?
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| Fig.2 |
Answer:
(a) The magnetic field is outward through the round coil and is decreasing, so the magnetic field due to the induced current must also point outward to oppse this decrease. Therefore the induced current is counterclockwise.
(b) ε = |dΦ/dt|= |d(BA)/dt| = A|dB/dt|
Then, the induced current is
I = ε/R = A|dB/dt|/R
I = π(0.0480 m)2(0.680 T/s)/0.160 Ω = 0.03076 A
So that, the rate at which electrical energy is being dissipated by the resistance of the loop is
P = I2R = (0.03076 A)(0.160 Ω)
P = 1.51 x 10-4 watt
Problem#3
A circular loop of wire with radius r = 0.0250 m and resistance R = 0.390 Ω is in a region of spatially uniform magnetic field, as shown in Fig. 3. The magnetic field is directed into the plane of the figure. At t = 0, B = 0. The magnetic field then begins increasing, with B(t) = (0.380 T/s3)t3 . What is the current in the loop (magnitude and direction) at the instant when B = 1.33 T?
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| Fig.3 |
Answer:
ε = |dΦ/dt|= |d(BA)/dt| = A|dB/dt|
B(t) = (0.380 T/s3)t3
for, B = 1.33 T
1.33 T = (0.380 T/s3)t3
t = 1.52 s
Then,
ε = A|dB/dt|
ε = πr2 x 3t2 x 0.380
ε = π(0.0250 m)2 x 3(1.52 s)2 x 0.380 = 5.15 x 10-3 V
the current in the loop is
I = ε/R = 5.15 x 10-3 V/0.390 Ω
I = 13.2 mA
B is increasing into page, then the current direction in the loop is counterclockwise.
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