Problem#1
A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.5 T is directed along the positive z-direction, which is upward. (a) If the loop is removed from the field region in a time interval of 2.0 ms, find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?
Answer:
We here to apply Farady’s Law. Let +z be the positive direction for A. Therefore, the initial flux is positive and the final flux is zero as
ε = –NΔΦB/Δt
with ΦB = BA cos θ, θi = 00, θf = 900 and N = 1.
When,
θi = 00, ΦB,i = 1.5 T cos 00 x π x (0.12 m)2 = 0.06788 Wb.
θf = 900, ΦB,f = 1.5 T x π x (0.12 m)2 cos 900 = 0 Wb.
(b) the average emf induced in the coil is
ε = –ΔΦB/Δt = (ΦB,f – ΦB,i)/Δt
ε = – (0 – 0.06788 Wb)/(0.0020 s) = +33.94 V
Since ε is positive and A is toward us, the induced current is counter-clockwise.
Problem#2
A coil 4.00 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B = (0.0120 T/s)t + (3.00 x 10–5 T/s4)t4. The coil is connected to a 600 Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time t = 5.00 s?
Answer:
(a) the magnitude of the induced emf in the coil as a function of time is
ε = –NdφB/dt = –NA (dB/dt)
with φ = BA cos θ, N = 500 turn and A = πr2
ε = –NA d(0.0120 T/s)t + (3.00 x 10–5 T/s4)t4)/dt
ε = –(500)[π(0.04 m)2] d(0.0120 T/s)t + (3.00 x 10–5 T/s4)t4)/dt
ε = 0.0302 V + (3.02 x 10–4 V/s3)t3
(b) at time t = 5.00 s,
ε = 0.0302 V + (3.02 x 10–4 V/s3)(5.00 s)3
ε = 0.0680 V
the current in the resistor is
I = ε/R = 0.0680 V/600 Ω = 1.13 x 10–4 A
Problem#3
The current in the long, straight wire AB shown in Fig. 1 is upward and is increasing steadily at a rate (a) At an instant when the current is i, what are the magnitude and direction of the field B at a distance r to the right of the wire? (b) What is the flux dφB through the narrow, shaded strip? (c) What is the total flux through the loop? (d) What is the induced emf in the loop? (e) Evaluate the numerical value of the induced emf if a = 12.0 cm, b = 24.0 cm, c = 36.0 cm and di/dt = 9.60 A/s.
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| Fig.1 |
Answer:
(a) At an instant when the current is i, what are the magnitude and direction of the field B at a distance r to the right of the wire? Consider the wire in isolation and use Ampere’s law,
that is
B = μ0i/2πr
and is into the paper.
(b) What is the flux dΦB through the narrow shaded strip?
dΦB = BLdr = (μ0i/2πr)Ldr
(c) What is the total flux ΦB through the loop? By integration
ΦB = ∫dΦB = (μ0i/2πr)L∫ab dr/r = μ0i/2πrL ln (b/a)
(d) What is the emf ε induced in the loop? Farday’s law states
ε = – dΦB/dt = – μ0i/2πrL[ln (b/a)]di/dt
(e) Evaluate the numerical value of E if a = 12.0cm, b = 36.0cm, L = 24.0cm and di/dt = 9.60A/s. Have the
ε = – μ0 x (0.24 A)/2π [ln (36 cm/12 cm)] x 9.6 A/s = 5.06 x 10–7 V
Problem#4
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 Fig. 2. The field is changing with time, according to B(t) = (1.4 T)e–0.057/s)t. (a) Find the emf induced in the loop as a function of time. (b) When is the induced emf equal to 1/10 of its initial value? (c) Find the direction of the current induced in the loop, as viewed from above the loop.
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| Fig.2 |
Answer:
(a) ΦB = B.A = BA cos θ
ΦB = (1.4 T)e–0.057/s)t [π x (0.75 m)2](sin 600)
ΦB = (2.14)e–0.057/s)t
the emf induced in the loop as a function of time
ε = –dΦB/dt
ε = –2.14 x (–0.057)e–0.057/s)t = (0.122 V)e–0.057/s)t
(b) if ε = ε0/10, then
0.122 V/10 = (0.122 V)e–0.057/s)t
1/10 = e–0.057/s)t
ln (0.1) = –(0.057/s)t
t = 40.4 s
(c) B is in the direction of A is positive. B is getting weaker, so the magnitude of the flux is decreasing and dΦB/dt < 0. Faraday’s law therefore says ε > 0. Since ε > 0, the induced current must flow counterclockwise as viewed from above.
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