Problem#1
A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 T/s. (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?Answer:
A changing magnetic flux through a coil induces an emf in that coil, which means that an electric field is induced in the material of the coil.
According to Faraday’s law, the induced electric field obeys the equation ∮ E.dL = -dφB/dt.
(a) For the magnitude of the induced electric field, Faraday’s law gives
E(2πr) = d(B.A)/dt = πr2 dB/dt
E = ½rdB/dt = 0.0225 m/2 x 0.250 T/s = 2.81 x 10-3 V/m
(b) The field points toward the south pole of the magnet and is decreasing, so the induced current is counterclockwise.
Problem#2
A long, straight solenoid with a cross-sectional area of 8.00 cm2 is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?
Answer:
Apply Faraday’s law in the form
The magnetic field of a large straight solenoid is B = μ0nI inside the solenoid and zero outside. , ΦB = BA where A is 8.00 cm2, the cross-sectional area of the long straight solenoid.
|εavr|= N|∆φB/∆t|
εavr = NA|(Bf – Bi)/∆t|
εavr = NAµ0nI/∆t
= 12 x (8.00 x 10-4 m2) x (4π x 10-7 T.m/A) x (9000/m)(0.350 A)/0.0400 s
εavr = 9.50 x 10-4 V
Problem#3
The magnetic field at all points within the colored circle shown in Fig. 1 has an initial magnitude of 0.750 T. (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of -0.0350 T/s. (a) What is the shape of the field lines of the induced electric field shown in Fig. 1, within the colored circle? (b) What are the magnitude and direction of this field at any point on the circular conducting ring with radius 0.100 m? (c) What is the current in the ring if its resistance is 4.00 Ω. (d) What is the emf between points a and b on the ring? (e) If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends?
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| Fig.1 |
(a) Because of the axial symmetry and the absence of any electric charge, the field lines are concentric circles.
(b) E is tangent to the ring. The direction of E (clockwise or counterclockwise) is the direction in which current will be induced in the ring. See Figure 2.
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| Fig.2 |
Use the sign convention for Faraday’s law to deduce this direction. Let A be into the paper. Then ΦB is positive. B decreasing then means dφB/dt is negative, so by ε = -dφB/dt. ε is positive and therefore clockwise. Thus E is clockwise around the ring. To calculate E apply ∮ E.dL = -dφB/dt to a circular path that coincides with the ring.
∮ E.dL = E(2πr) and
|-dφB/dt| = πr2 |dB/dt|
Then,
E(2πr) = πr2 |dB/dt|
E = ½ r|dB/dt|
E = ½ x 0.100 m x 0.0350 T/s = 1.75 x 10-3 V/m
(c) The induced emf has magnitude
ε = ∮ E.dL = E(2πr)
ε = 1.75 x 10-3 V/m (2π x 0.100 m) = 1.100 x 10-3 V
Then,
I = ε/R = 1.100 x 10-3 V/4.00Ω = 2.75 x 10-4 A
(d) Points a and b are separated by a distance around the ring of πr so
ε = E(πr) = (1.75 10 V/m)(π)(0.100 m) = 5.50 x 10-4 V
(e) The ends are separated by a distance around the ring of 2πr so ε = 1.10 x 10-3 V as calculated in part (c).
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