Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area $1.20 \times 10^{- 3} m^{2}$ . The two ends of the wire are connected to a resistor. The total resistance in the circuit is $13.0 Ω$ . If an externally applied uniform longitudinal magnetic field in the core changes from 1.60 T in one direction to 1.60 T in the opposite direction, how much charge flows through a point in the circuit during the change?

The charge flowing through a point in the circuit during the change in magnetic field is, $q (t) = 2.95 \times 10^{- 2} C$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Cross-sectional area, $A = 1.20 \times 10^{- 3} m^{2}$
Resistance, $R = 13.0 Ω$
Magnetic field at core, B (0) = 1.60 T
Magnetic field at other end in opposite direction, B (t) = (-1.60 T)

Step 2: Determining the concept

The magnetic flux $φ_{B}$ through area A in a magnetic field is given by Faraday’s law of induction. If the magnetic field $\vec{B}$ is perpendicular to the area A, then substitute the given values in the formula to find the charge.

Faraday's law of electromagnetic induction states, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

$e m f = - N . \frac{d φ}{d t}$

Where, $d Φ$ is magnetic flux, N is number of turns, dt is time.

Step 3: Determining the charge flowing through a point in the circuit during the change in magnetic field

According to Faraday’s law,

$e m f = - N . \frac{d φ}{d t}$

From Ohm’s law, emf = i.R and i = dq/dt

$∴ i . R = - N . \frac{d φ}{d t}$

$\frac{d q}{d t} = - N . \frac{d φ}{d t}$

Integrating this with respect to time,

$q {(t)}_{0}^{t} = - N φ {(t)}_{0}^{t}$

To calculate the charge flowing through the point,

$q (t) = \frac{N}{R} [φ_{B} (0) - φ_{B} (t)]$

According to Faraday’s law, if the magnetic field $\vec{B}$ is perpendicular to the area A, then,

$φ_{B} = B A$

Therefore,

$q (t) = \frac{NA}{R} [B (0) - B (t)] q (t) = \frac{100 \times (1.20 \times 10^{- 3} m^{2})}{(13.0 Ω)} [16.0 T - (- 1.60 T)] q (t) = 100 \times (0.0923 \times 10^{- 3} m^{2}) \times (3.20) q (t) = 2.95 \times 10^{- 2} C$

Hence, the charge flowing through a point in the circuit during the change in magnetic field is, $q (t) = 2.95 \times 10^{- 2} C$

Therefore, the charge flow through a point in the circuit during the change in the magnetic field can be found by using Faraday’s law.

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