JEE MAIN - Physics (2024 - 8th April Morning Shift - No. 24)
A square loop PQRS having 10 turns, area $$3.6 \times 10^{-3} \mathrm{~m}^2$$ and resistance $$100 \Omega$$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $$\mathrm{B}=0.5 \mathrm{~T}$$ as shown. Work done in pulling the loop out of the field in $$1.0 \mathrm{~s}$$ is _________ $$\times 10^{-6} \mathrm{~J}$$.
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Answer
3
Explanation
$$\begin{aligned}
& A = 36 \times 10^{-4} \mathrm{~m}^2 \\
& I= 6 \times 10^{-2} \mathrm{~m} \\
& =6 \mathrm{~cm} \\
V & =\frac{6 \mathrm{~cm}}{1 \mathrm{sec}}=6 \mathrm{~cm} / \mathrm{s} \\
\varepsilon & =B / \mathrm{vn}^2=0.5 \times \frac{6}{100} \times \frac{6}{100} \\
& =18 \times 10^{-4} \mathrm{~V} \\
E & =\frac{n^2 \varepsilon^2}{R} t=100 \times \frac{18 \times 18 \times 10^{-4} \times 10^{-4}}{10^2} \times 1 \\
& =324 \times 10^{-10} \times 10^2 \\
& =3.24 \times 10^{-6}
\end{aligned}$$
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