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Found in: Page 896

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 ${{\mathbf{B}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{}}{\mathbf{T}}$and the horizontal scale is set by ${{\mathbf{t}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbit{s}}$ A circular conducting loop of area ${\mathbf{8}}{\mathbf{.}}{\mathbf{0}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{4}}{{\mathbf{m}}}^{{\mathbf{2}}}$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 ${{\mathbit{q}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbit{s}}$ and the horizontal axis scale again set by localid="1661854094654" ${{\mathbf{t}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{s}}$. What is the loop’s resistance?

Resistance of the loop is, R = $R=0.0012\Omega$$0.0012\Omega$

See the step by step solution

## Step 1: Given

1. Cross-sectional area $\mathrm{A}=8×{10}^{-4}{\mathrm{m}}^{2}$
2. Magnetic field on vertical axis, ${\mathrm{B}}_{\mathrm{s}}=9.0\mathrm{mT}=9×{10}^{-3}\mathrm{T}$
3. Horizontal axis, ${\mathrm{t}}_{\mathrm{s}}=3.0\mathrm{s}$
4. Vertical axis, ${\mathrm{q}}_{\mathrm{s}}=6\mathrm{mC}=6×{10}^{-3}\mathrm{C}$

## 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:

$\epsilon =\frac{d{\phi }_{B}}{dt}\phantom{\rule{0ex}{0ex}}i=\frac{dq}{dt}\phantom{\rule{0ex}{0ex}}R=\frac{\left|\epsilon \right|}{i}$

Where, ${\Phi }_{B}$ 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,

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

$\epsilon =-\frac{d{\phi }_{B}}{dt}$

As magnetic field is perpendicular to the area vector,

${\phi }_{B}=BA$

Therefore,

role="math" localid="1661853786477" $\epsilon =BA\phantom{\rule{0ex}{0ex}}\epsilon =\frac{d\left(BA\right)}{dt}\phantom{\rule{0ex}{0ex}}\epsilon =-A\frac{dB}{dt}$

Substituting the values,

$\mathrm{\epsilon }=-\left(8.0\mathrm{x}{10}^{-4}\right)\left(0.003\right)\phantom{\rule{0ex}{0ex}}\mathrm{\epsilon }=-0.024×{10}^{-4}\mathrm{V}$

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

The slope of the graph gives dq/dt.

Therefore,

$\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{6×{10}^{-3}\mathrm{C}}{3.0\mathrm{s}}\phantom{\rule{0ex}{0ex}}\mathrm{i}=0.002\mathrm{A}$

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

$R=\frac{\left|\epsilon \right|}{i}$

Substituting the values,

$R=\frac{\left|-0.024×{10}^{-4}\right|}{0.002}\phantom{\rule{0ex}{0ex}}R=0.0012\Omega$

Hence, resistance of the loop is, $R=0.0012\Omega$

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