Problem#1
In Fig. 1 the loop is being pulled to the right at constant speed v. A constant current I flows in the long wire, in the direction shown. (a) Calculate the magnitude of the net emf ε induced in the loop. Do this two ways: (i) by using Faraday’s law of induction and (ii) by looking at the emf induced in each segment of the loop due to its motion. (b) Find the direction (clockwise or counterclockwise) of the current induced in the loop. Do this two ways: (i) using Lenz’s law and (ii) using the magnetic force on charges in the loop. (c) Check your answer for the emf in part (a) in the following special cases to see whether it is physically reasonable: (i) The loop is stationary; (ii) the loop is very thin, so a → 0 (iii) the loop gets very far from the wire.
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| Fig.1 |
Answer:
(a) (i) │ε│ = │dΦB/dt│ is the flux is changing because the magnitude of the magnetic field of the wire decreases with distance from the wire. Find the flux through a narrow strip of area and integrate over the loop to find the total flux.
Consider a narrow strip of width dx and a distance x from the long wire, as shown in Figure 2a. The magnetic field of the wire at the strip is
B =μ0I/2πx
The flux through the strip is
dΦB = Bbdx = (μ0Ib/2π)(dx/x)
Then, the total flux through the loop is
ΦB = ∫rr+a (μ0Ib/2π)(dx/x)
ΦB = (μ0Ib/2π)ln(1 + a/r)
dΦB/dt = (dΦB/dr)(dr/dt)
│ε│ = μ0Iabv/[2πr(r + a)]
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| Fig.2 |
(ii) ε = BvL for a bar of length l moving at speed v perpendicular to a magnetic field B. Calculate the induced emf in each side of the loop, and combine the emfs according to their polarity. The four segments of the loop are shown in Figure 2b.
The emf in each side of the loop is
ε1 = μ0Ibv/2πr
ε2 = μ0Ibv/[2πr(r + a)]
ε3 = ε4 = 0
Both emfs ε1 and ε2 are directed toward the top of the loop so oppose each other. The net emf is
ε = ε1 – ε2
ε = μ0Iabv/[2πr(r + a)]
This expression agrees with what was obtained in (i) using Faraday’s law.
(b) (i) The flux of the induced current opposes the change in flux.
B is ⊗ ΦB is decreasing, so the flux Φind of the induced current is ⊗ and the current is clockwise.
(ii) Use the right-hand rule to find the force on the positive charges in each side of the loop. The forces on positive charges in segments 1 and 2 of the loop are shown in Figure 3.
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| Fig.3 |
Problem#2
Suppose the loop in Fig. 4 is (a) rotated about the y-axis; (b) rotated about the x-axis; (c) rotated about an edge parallel to the z-axis. What is the maximum induced emf in each case if A = 600 cm2, ω = 35.0 rad/s, and B = 0.450 T?
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| Fig.4 |
Answer:
Apply Faraday’s law.
(a) Rotating about the y-axis: the flux is given by ΦB = BAcosφ and
εmax = ωBA = (35.0 rad/s)(0.450 T)(6.00 x 10-2 m) = 0.945 V.
(b) Rotating about the x-axis: dΦB/dt = 0 and ε = 0.
(c) Rotating about the z-axis: the flux is given by ΦB = BAcosφ and
εmax = ωBA = (35.0 rad/s)(0.450 T)(6.00 x 10-2 m) = 0.945 V.
The maximum emf is the same if the loop is rotated about an edge parallel to the z-axis as it is when it is rotated about the z-axis.
Problem#3
As a new electrical engineer for the local power company, you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. Besides plenty of wire, you have two strong magnets that can produce a constant uniform magnetic field of 1.5 T over a square area of 10.0 cm on a side when they are 12.0 cm apart. The basic design should consist of a square coil turning in the uniform magnetic field. To have an acceptable coil resistance, the coil can have at most 400 loops. What is the minimum rotation rate (in rpm) of the coil so it will produce the required voltage?
Answer:
For the minimum ω, let the rotating loop have an area equal to the area of the uniform magnetic field, so A = (0.100 m)2 and because εmax = NωBA, then the minimum rotation rate (in rpm) of the coil so it will produce the required voltage is
ω = εmax/NBA
= 120 V/(400 x 1.5 T x 0.01 m2)
= (20 rad/s)(1 rev/2π rad)(60 s/1 min)
ω = 190 rpm.
Problem#4
A flexible circular loop 6.50 cm in diameter lies in a magnetic field with magnitude 1.35 T, directed into the plane of the page as shown in Fig. 5. The loop is pulled at the points indicated by the arrows, forming a loop of zero area in 0.250 s. (a) Find the average induced emf in the circuit. (b) What is the direction of the current in R: from a to b or from b to a? Explain your reasoning.
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| Fig.5 |
Answer:
(a) Apply Faraday’s law in the form εav = –N∆ΦB/∆t to calculate the average emf. Apply Lenz’s law to calculate the direction of the induced current. With the flux changes because the area of the loop changes ΦB = BA.
εav = │∆ΦB/∆t│ = B│∆A/∆t│
εav = Bπr2/∆t
εav = π x 1.35 T x (0.0650 m/2)2/0.250 s
εav = 17.9 mV
(b) Since the magnetic field is directed into the page and the magnitude of the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore the current flows from point a through the resistor to point b.
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