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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
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The correct answer is:
$1.0 \times 10^{-3} \mathrm{~J}$
Given: area of square loop $=25 \mathrm{~cm}$
$\begin{array}{ll}
\therefore \quad l=\sqrt{25}=5 \mathrm{~cm} \Rightarrow 0.05 \mathrm{~m} \\
\quad & \mathrm{R}=10 \Omega, \mathrm{t}=1 \mathrm{sec}, \mathrm{B}=40 \mathrm{~T} \\
\therefore \quad & \text { Velocity } \mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s} \\
& \text { Motional emf } \varepsilon_{\max }=\mathrm{B} / \mathrm{v}
\\
\therefore \quad & \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}} \\
\therefore \quad & \mathrm{I}=\frac{40 \times 0.05 \times 0.05}{10}=0.01 \mathrm{~A}
\end{array}$
We know, Force acting on loop
$\begin{aligned}
\therefore\mathrm{F}=\mathrm{BI} l & =40 \times 0.01 \times 0.05 \\
& =0.02 \mathrm{~N} \\
\text { Using W }=\mathrm{F} . \mathrm{s}, & \\
\text { Work done } \mathrm{W} & =\mathrm{BI} l \times l \\
& =0.02 \times 0.05 \\
& =1 \times 10^{-3} \mathrm{~J}
\end{aligned}$
$\begin{array}{ll}
\therefore \quad l=\sqrt{25}=5 \mathrm{~cm} \Rightarrow 0.05 \mathrm{~m} \\
\quad & \mathrm{R}=10 \Omega, \mathrm{t}=1 \mathrm{sec}, \mathrm{B}=40 \mathrm{~T} \\
\therefore \quad & \text { Velocity } \mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s} \\
& \text { Motional emf } \varepsilon_{\max }=\mathrm{B} / \mathrm{v}
\\
\therefore \quad & \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}} \\
\therefore \quad & \mathrm{I}=\frac{40 \times 0.05 \times 0.05}{10}=0.01 \mathrm{~A}
\end{array}$
We know, Force acting on loop
$\begin{aligned}
\therefore\mathrm{F}=\mathrm{BI} l & =40 \times 0.01 \times 0.05 \\
& =0.02 \mathrm{~N} \\
\text { Using W }=\mathrm{F} . \mathrm{s}, & \\
\text { Work done } \mathrm{W} & =\mathrm{BI} l \times l \\
& =0.02 \times 0.05 \\
& =1 \times 10^{-3} \mathrm{~J}
\end{aligned}$
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