Question 283: Electromagnetic Induction

Question

A square loop of area $$25 \mathrm{~cm}^2$$ has a resistance of $$10 \Omega$$. This loop is placed in a uniform magnetic field of magnitude $$40 \mathrm{~T}$$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in one second, will be

Accepted Answer

Given: area of square loop $$=25 \mathrm{~cm}$$
$$\begin{array}{ll}
	herefore \quad l=\sqrt{25}=5 \mathrm{~cm} \Rightarrow 0.05 \mathrm{~m} \
\quad \mathrm{R}=10 \Omega, \mathrm{t}=1 \mathrm{sec}, \mathrm{B}=40 \mathrm{~T} \
	herefore \quad 	ext { Velocity } \mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s} \
\quad 	ext { Motional emf } \varepsilon_{\max }=\mathrm{B} / \mathrm{v} \
	herefore \quad \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}} \
	herefore \quad \
\quad \mathrm{I}=\frac{40 	imes 0.05 	imes 0.05}{10}=0.01 \mathrm{~A}
\end{array}$$
We know, Force acting on loop
$$\begin{aligned}
& 	herefore \quad \mathrm{F}=\mathrm{BI} l=40 	imes 0.01 	imes 0.05 \
& =0.02 \mathrm{~N} \
& 	ext { Using } \mathrm{W}=\mathrm{F} . \mathrm{s} 	ext {, } \
& 	ext { Work done } \mathrm{W}=\mathrm{BI} l 	imes l \
& =0.02 	imes 0.05 \
& =1 	imes 10^{-3} \mathrm{~J} \
\end{aligned}$$

Answer

$$1 	imes 10^{-4} \mathrm{~J}$$

Answer

$$1.0 	imes 10^{-3} \mathrm{~J}$$

Answer

$$5 	imes 10^{-3} \mathrm{~J}$$

Answer

$$2.5 	imes 10^{-3} \mathrm{~J}$$

Electromagnetic Induction

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