Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 31, Problem 11P

(a)

To determine

The induced current in the ring.

(a)

Expert Solution
Check Mark

Answer to Problem 11P

The induced current in the ring is 1.60A in counterclockwise direction.

Explanation of Solution

Write the expression to calculate the magnetic flux through the loop.

    ΦB=BA2                                                                                                                  (I)

Here, ΦB is the magnetic flux, B is the magnetic field, A is the area

Write the expression for the induced emf by Faraday law.

    E=ddt(ΦB)                                                                                                              (II)

Here, E is the induced emf.

Write the expression for the area.

    A=πr22                                                                                                                   (III)

Here, r2 is the radius of the solenoid.

Write the expression for the magnetic field.

    B=μ0NI                                                                                                                (IV)

Here, B is the magnetic field, N is the number of turns, I is the current an μ0 is the permeability.

Substitute BA2 for ΦB, πr22 for A and μ0NI for B in equation (II).

    E=ddt((μ0NI)(πr22)2)=((μ0N)(πr22)2)dIdt                                                                                           (V)

Write the expression for the induced current.

    i=ER                                                                                                                         (VI)

Substitute ((μ0N)(πr22)2)dIdt for E in equation (VI),

    i=1R(((μ0N)(πr22)2)dIdt)                                                                                   (VII)

Conclusion:

Substitute 3.00×104Ω for R, 3.00cm for r2, 270A/s for dIdt, 4π×107T-m/A for μ0 and 1000turns/m for N in equation (VII) to solve for i.

    i=((13.00×104Ω)(270A/s)((4π×107T-m/A)(1000turns/m)(π(3.00cm×102m1cm)2)2))=1.598A1.60A

The direction of the current is counter clockwise.

Therefore, the induced current in the ring is 1.60A in counterclockwise direction.

(b)

To determine

The magnitude of the magnetic field at the center of the ring

(b)

Expert Solution
Check Mark

Answer to Problem 11P

The magnitude of the magnetic field at the center of the ring is 20.1μT.

Explanation of Solution

Write the expression to calculate the magnetic field at the center of the ring.

    B=μ0i2r1                                                                                                                 (VII)

Here, B is the magnetic field at the center of the ring and r1 is the radius of the ring.

Conclusion:

Substitute 5.00cm for r1, 1.60A for i and 4π×107T-m/A for μ0 in equation (VIII) to solve for B.

    B=(4π×107T-m/A)(1.60A)2(5.00cm×102m1cm)=20.1×106T×106μT1T=20.1μT

Therefore, the magnitude of the magnetic field at the center of the ring is 20.1μT.

(c)

To determine

The direction of the magnetic field at the center of the ring.

(c)

Expert Solution
Check Mark

Answer to Problem 11P

The direction of the magnetic field at the center of the ring is to the left.

Explanation of Solution

The magnetic field of the solenoid points towards the right but the induced field opposes the original field. So, the magnetic field at the center of the ring will be directed towards the left.

Conclusion:

Therefore, the direction of the magnetic field at the center of the ring is to the left.

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Chapter 31 Solutions

Physics for Scientists and Engineers With Modern Physics

Ch. 31 - Prob. 6OQCh. 31 - Prob. 7OQCh. 31 - Prob. 8OQCh. 31 - Prob. 9OQCh. 31 - Prob. 10OQCh. 31 - Prob. 11OQCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQCh. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 10CQCh. 31 - Prob. 1PCh. 31 - Prob. 2PCh. 31 - Prob. 3PCh. 31 - Prob. 4PCh. 31 - Prob. 5PCh. 31 - Prob. 6PCh. 31 - Prob. 7PCh. 31 - Prob. 8PCh. 31 - Prob. 9PCh. 31 - Scientific work is currently under way to...Ch. 31 - Prob. 11PCh. 31 - Prob. 12PCh. 31 - Prob. 13PCh. 31 - Prob. 14PCh. 31 - Prob. 15PCh. 31 - Prob. 16PCh. 31 - A coil formed by wrapping 50 turns of wire in the...Ch. 31 - Prob. 18PCh. 31 - Prob. 19PCh. 31 - Prob. 20PCh. 31 - Prob. 21PCh. 31 - Prob. 22PCh. 31 - Prob. 23PCh. 31 - A small airplane with a wingspan of 14.0 m is...Ch. 31 - A 2.00-m length of wire is held in an eastwest...Ch. 31 - Prob. 26PCh. 31 - Prob. 27PCh. 31 - Prob. 28PCh. 31 - Prob. 29PCh. 31 - Prob. 30PCh. 31 - Prob. 31PCh. 31 - Prob. 32PCh. 31 - Prob. 33PCh. 31 - Prob. 34PCh. 31 - Prob. 35PCh. 31 - Prob. 36PCh. 31 - Prob. 37PCh. 31 - Prob. 38PCh. 31 - Prob. 39PCh. 31 - Prob. 40PCh. 31 - Prob. 41PCh. 31 - Prob. 42PCh. 31 - Prob. 43PCh. 31 - Prob. 44PCh. 31 - Prob. 45PCh. 31 - Prob. 46PCh. 31 - Prob. 47PCh. 31 - Prob. 48PCh. 31 - The rotating loop in an AC generator is a square...Ch. 31 - Prob. 50PCh. 31 - Prob. 51APCh. 31 - Prob. 52APCh. 31 - Prob. 53APCh. 31 - Prob. 54APCh. 31 - Prob. 55APCh. 31 - Prob. 56APCh. 31 - Prob. 57APCh. 31 - Prob. 58APCh. 31 - Prob. 59APCh. 31 - Prob. 60APCh. 31 - Prob. 61APCh. 31 - Prob. 62APCh. 31 - Prob. 63APCh. 31 - Prob. 64APCh. 31 - Prob. 65APCh. 31 - Prob. 66APCh. 31 - Prob. 67APCh. 31 - A conducting rod moves with a constant velocity in...Ch. 31 - Prob. 69APCh. 31 - Prob. 70APCh. 31 - Prob. 71APCh. 31 - Prob. 72APCh. 31 - Prob. 73APCh. 31 - Prob. 74APCh. 31 - Prob. 75APCh. 31 - Prob. 76APCh. 31 - Prob. 77APCh. 31 - Prob. 78APCh. 31 - Prob. 79CPCh. 31 - Prob. 80CPCh. 31 - Prob. 81CPCh. 31 - Prob. 82CPCh. 31 - Prob. 83CP
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