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  • A square loop of wire with side length 10 cm is placed at angle of 45° with a magnetic field that changes uniformly from 0.1 T to zero in 0.7 s. The induced current in the loop (its resistance

A square loop of wire with side length 10 cm is placed at angle of 45° with a magnetic field that changes uniformly from 0.1 T to zero in 0.7 s. The induced current in the loop (its resistance is1\Omega ) is:

Option: 1

1.0 mA


Option: 2

2.5 mA


Option: 3

3.5 mA


Option: 4

4.0 mA


Answers (1)

best_answer

Area of square loop, \mathrm{S=10 cm\times10 cm}

\mathrm{\Rightarrow \mathrm{S}=100 \mathrm{~cm}^2=100 \times 10^{-4} \mathrm{~m}^2=10^{-2} \mathrm{~m}^2}

Initial magnetic flux linked with loop, \mathrm{\phi_{\mathrm{B}_1}=\mathrm{B}_1 \mathrm{~S} \cos \phi=0.1 \times 10^{-2} \times \cos 45^{\circ}=\frac{0.1 \times 10^{-2} \times 1}{\sqrt{2}}=\frac{10^{-3}}{\sqrt{2}} \mathrm{~Wb}}

Final magnetic flux linked with loop, \mathrm{\phi_{\mathrm{B}_2}=0 \mathrm{~Wb} \quad\left[\because \mathrm{B}_2=0\right]}

The induced emf in the loop, \mathrm{\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\left(\phi_2-\phi_1\right)}{\mathrm{t}}=-\frac{\left(0-\frac{10^{-3}}{\sqrt{2}}\right)}{0.7}=\frac{10^{-3}}{0.7 \times \sqrt{2}} \cong 10^{-3} \mathrm{~V}}

The induced current in the loop,\mathrm{I=\frac{\varepsilon}{R}=\frac{10^{-3} \mathrm{v}}{1 \Omega}=10^{-3} \mathrm{~A}=1.0 \mathrm{~mA}}

Posted by

shivangi.bhatnagar

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