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A wire of length $1 \mathrm{~m}$ is moving at a speed of $2 \mathrm{~m} / \mathrm{s}$ perpendicular homogenous magnetic field of $0.5 \mathrm{~T}$. The ends of the wire are joined to resistance $6 \Omega$. The rate at which work is being done to keep the wire moving at that speed is
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The correct answer is:
$\frac{1}{6} \mathrm{~W}$
Emf induced e $=\mathrm{B} \ell \mathrm{v}=0.5 \times 1 \times 2=1 \mathrm{~V}$
Rate of doing work $=$ Power
$$
\mathrm{P}=\frac{\mathrm{e}^2}{\mathrm{R}}=\frac{(1)^2}{6}=\frac{1}{6} \mathrm{~W}
$$
Rate of doing work $=$ Power
$$
\mathrm{P}=\frac{\mathrm{e}^2}{\mathrm{R}}=\frac{(1)^2}{6}=\frac{1}{6} \mathrm{~W}
$$
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