Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Figure (a), a uniform magnetic field increases in magnitude with time t as given by Figure (b), where the vertical axis scale is set by $B_{s} = 9.0 m T$ and the horizontal scale is set by $t_{s} = 3.0 s$ A circular conducting loop of area $8.0 \times 10^{- 4} m^{2}$ lies in the field, in the plane of the page. The amount of charge q passing point A on the loop is given in Figure (c) as a function of t, with the vertical axis scale set by $q_{s} = 3.0 s$ and the horizontal axis scale again set by localid="1661854094654" $t_{s} = 3.0 s$ . What is the loop’s resistance?

Resistance of the loop is, R = $R = 0.0012 Ω$ $0.0012 Ω$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Cross-sectional area $A = 8 \times 10^{- 4} m^{2}$
Magnetic field on vertical axis, $B_{s} = 9.0 mT = 9 \times 10^{- 3} T$
Horizontal axis, $t_{s} = 3.0 s$
Vertical axis, $q_{s} = 6 mC = 6 \times 10^{- 3} C$

Step 2: Determining the concept

Calculate the induced emf by using the slope of the line form Fig. 30-42(b) in Faraday’s law. Find the current by finding the slope of the line from Fig.30-42 (c) to find the resistance of the loop by using Ohm’s law.

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

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:

$ε = \frac{d φ_{B}}{d t} i = \frac{d q}{d t} R = \frac{|ε|}{i}$

Where, $Φ_{B}$ is magnetic flux, dt is time, dq is charge, i is current, R is resistance, 𝜀 is emf,

Step 3: Determining the resistance of the loop

First, find the emf by using Faraday’s law.

From Fig.30 - 42 (b) the slope of the line i.e. dB/dt can be found as,

$\frac{dB}{dt} = \frac{9 \times 10^{- 3} T}{3.0 s} = 0.003 T / s$

Faraday’s law is given as,

$ε = - \frac{d φ_{B}}{d t}$

As magnetic field is perpendicular to the area vector,

$φ_{B} = B A$

Therefore,

role="math" localid="1661853786477" $ε = B A ε = \frac{d (B A)}{d t} ε = - A \frac{d B}{d t}$

Substituting the values,

$ε = - (8.0 x 10^{- 4}) (0.003) ε = - 0.024 \times 10^{- 4} V$

Now, the current can be calculated from Fig. 30- 42 (c).

The slope of the graph gives dq/dt.

Therefore,

$i = \frac{dq}{dt} = \frac{6 \times 10^{- 3} C}{3.0 s} i = 0.002 A$

Now, relate the induced emf to resistance and current by using Ohm’s law.

$R = \frac{|ε|}{i}$

Substituting the values,

$R = \frac{|- 0.024 \times 10^{- 4}|}{0.002} R = 0.0012 Ω$

Hence, resistance of the loop is, $R = 0.0012 Ω$

Therefore, we calculated the resistance of the loop by using the slopes from the graph, Faraday’s law, and Ohm’s law.

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