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Q14P

Expert-verifiedFound in: Page 896

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 ${{\mathbf{B}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{}}{\mathbf{T}}$**

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

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

**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,

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,

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

Faraday’s law is given as,

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

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,

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

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

The slope of the graph gives dq/dt.

Therefore,

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

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\times {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.

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