Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 23, Problem 12P

A piece of insulated wire is shaped into a figure eight as shown in Figure P23.12. For simplicity, model the two halves of the figure eight as circles. The radius of the upper circle is 5.00 cm and that of the lower circle is 9.00 cm. The wire has a uniform resistance per unit length of 3.00 Ω/m. A uniform magnetic field is applied perpendicular to the plane of the two circles, in the direction shown. The magnetic field is increasing at a constant rate of 2.00 T/s. Find (a) the magnitude and (b) the direction of the induced current in the wire.

Chapter 23, Problem 12P, A piece of insulated wire is shaped into a figure eight as shown in Figure P23.12. For simplicity,

Figure P23.12

(a)

Expert Solution
Check Mark
To determine

The magnitude of the induced current

Answer to Problem 12P

The induced current has a magnitude of 0.0133A.

Explanation of Solution

The magnetic field in which the wire is placed is increasing and points into the page. This increasing flux is opposed by a magnetic field out of the page due to the counterclockwise current as a result of induced emf in the upper loop. Similarly, the lower loop also develops an induced emf.

The emf generated in the upper and the lower loops are in opposite directions as the two loops cross over. Write the equation for the net emf generated which is the difference between the emf generated in the two loops.

    εnet=ε2ε1        (I)

Here, εnet is the net emf generated, ε2 is the emf in the lower loop and ε1 is the emf in the lower loop. Substitute A2(dB/dt) for ε2 and A1(dB/dt) for ε1 in equation (I).

    εnet=A2(dBdt)A1(dBdt)        (II)

Here, A2 is the area of the lower loop, A1 is the area of the upper loop and (dB/dt) is the change in magnetic field. Substitute πr22 for A2 and πr12 for A1 in equation (II).

    εnet=πr22(dBdt)πr12(dBdt)=πdBdt(r22r12)        (III)

Here, r2 is the radius of the lower loop and r1 is the radius of the upper loop. Write the equation for the induced current in the wire.

    I=εnetR        (IV)

Here, I is the current induced in the wire, εnet is the net emf generated and R is the resistance of the wire. Substitute equation (III) in equation (IV).

    I=πdBdt(r22r12)R        (V)

Multiply and divide the denominator of equation (V) with the length of the wire l.

    I=πdBdt(r22r12)(Rl)l        (VI)

The length of the wire is the sum of the circumferences of the lower loop and the upper loop. The circumference of the lower loop is 2πr2 and the circumference of the upper loop is 2πr1. Use this to rewrite the equation (VI) and solve.

    I=πdBdt(r22r12)(Rl)(2πr2+2πr1)=πdBdt(r2r1)(r2+r1)2π(Rl)(r2+r1)=dBdt(r2r1)2(Rl)

Conclusion:

Substitute 2.00T/s for dB/dt, 9.00cm for r2, 5.00cm for r1 and 3.00Ω/m for R/l.

    I=2.00T/s(9.00cm5.00cm)2(3.00Ω/m)=0.0133A

Therefore, the induced current has a magnitude of 0.0133A.

(b)

Expert Solution
Check Mark
To determine

The direction of the induced current

Answer to Problem 12P

The induced current is counterclockwise in the lower loop and clockwise in the upper loop.

Explanation of Solution

The lower loop has a larger area and hence the emf. As a result, the change in magnetic flux is also larger in the lower loop.

This increasing flux is opposed by a magnetic field out of the page due to the counterclockwise current as a result of induced emf in the upper loop.

Conclusion:

Therefore, the induced current is counterclockwise in the lower loop and clockwise in the upper loop.

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The figure below shows a piece of insulated wire formed into the shape of a figure eight. You may consider the two loops of the figure eight to be circles, where the upper loop's radius is 3.00 cm and the lower loop's radius is 8.00 cm. The wire has a uniform resistance per unit length of 9.00 Ω/m. The wire lies in a plane that is perpendicular to a uniform magnetic field directed into the page, the magnitude of which is increasing at a constant rate of 1.70 T/s. What is the magnitude of the induced current in the wire (in A)?
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Chapter 23 Solutions

Principles of Physics: A Calculus-Based Text

Ch. 23 - Prob. 2OQCh. 23 - Prob. 3OQCh. 23 - A circular loop of wire with a radius of 4.0 cm is...Ch. 23 - A rectangular conducting loop is placed near a...Ch. 23 - Prob. 6OQCh. 23 - Prob. 7OQCh. 23 - Prob. 8OQCh. 23 - A square, flat loop of wire is pulled at constant...Ch. 23 - The bar in Figure OQ23.10 moves on rails to the...Ch. 23 - Prob. 11OQCh. 23 - Prob. 12OQCh. 23 - A bar magnet is held in a vertical orientation...Ch. 23 - Prob. 14OQCh. 23 - Two coils are placed near each other as shown in...Ch. 23 - A circuit consists of a conducting movable bar and...Ch. 23 - Prob. 17OQCh. 23 - Prob. 1CQCh. 23 - Prob. 2CQCh. 23 - Prob. 3CQCh. 23 - Prob. 4CQCh. 23 - Prob. 5CQCh. 23 - Prob. 6CQCh. 23 - Prob. 7CQCh. 23 - Prob. 8CQCh. 23 - Prob. 9CQCh. 23 - Prob. 10CQCh. 23 - Prob. 11CQCh. 23 - Prob. 12CQCh. 23 - Prob. 13CQCh. 23 - Prob. 14CQCh. 23 - Prob. 15CQCh. 23 - Prob. 16CQCh. 23 - Prob. 1PCh. 23 - An instrument based on induced emf has been used...Ch. 23 - A flat loop of wire consisting of a single turn of...Ch. 23 - Prob. 4PCh. 23 - Prob. 5PCh. 23 - Prob. 6PCh. 23 - A loop of wire in the shape of a rectangle of...Ch. 23 - When a wire carries an AC current with a known...Ch. 23 - Prob. 9PCh. 23 - Prob. 10PCh. 23 - Prob. 11PCh. 23 - A piece of insulated wire is shaped into a figure...Ch. 23 - A coil of 15 turns and radius 10.0 cm surrounds a...Ch. 23 - Prob. 14PCh. 23 - Figure P23.15 shows a top view of a bar that can...Ch. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - A metal rod of mass m slides without friction...Ch. 23 - Review. After removing one string while...Ch. 23 - Prob. 20PCh. 23 - The homopolar generator, also called the Faraday...Ch. 23 - Prob. 22PCh. 23 - A long solenoid, with its axis along the x axis,...Ch. 23 - Prob. 24PCh. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - A coil of area 0.100 m2 is rotating at 60.0 rev/s...Ch. 23 - A magnetic field directed into the page changes...Ch. 23 - Within the green dashed circle shown in Figure...Ch. 23 - Prob. 30PCh. 23 - Prob. 31PCh. 23 - Prob. 32PCh. 23 - Prob. 33PCh. 23 - Prob. 34PCh. 23 - Prob. 35PCh. 23 - Prob. 36PCh. 23 - Prob. 37PCh. 23 - Prob. 38PCh. 23 - Prob. 39PCh. 23 - Prob. 40PCh. 23 - Prob. 41PCh. 23 - Prob. 42PCh. 23 - Prob. 43PCh. 23 - Prob. 44PCh. 23 - Prob. 45PCh. 23 - Prob. 46PCh. 23 - Prob. 47PCh. 23 - Prob. 48PCh. 23 - Prob. 49PCh. 23 - Prob. 50PCh. 23 - Prob. 51PCh. 23 - Prob. 52PCh. 23 - Prob. 53PCh. 23 - Prob. 54PCh. 23 - Prob. 55PCh. 23 - Prob. 56PCh. 23 - Prob. 57PCh. 23 - Figure P23.58 is a graph of the induced emf versus...Ch. 23 - Prob. 59PCh. 23 - Prob. 60PCh. 23 - The magnetic flux through a metal ring varies with...Ch. 23 - Prob. 62PCh. 23 - Prob. 63PCh. 23 - Prob. 64PCh. 23 - Prob. 65PCh. 23 - Prob. 66PCh. 23 - Prob. 67PCh. 23 - Prob. 68PCh. 23 - Prob. 69PCh. 23 - Prob. 70PCh. 23 - Prob. 71PCh. 23 - Prob. 72PCh. 23 - Review. The use of superconductors has been...Ch. 23 - Prob. 74PCh. 23 - Prob. 75P
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