A wire of length $1 \mathrm{~m}$ is moving at a speed of $2 \mathrm{~m} / \mathrm{s}$ perpendicular…

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
  1. $\frac{1}{3} \mathrm{~W}$
  2. $\frac{1}{6} \mathrm{~W}$
  3. $\frac{1}{12} \mathrm{~W}$
  4. $1 \mathrm{~W}$

Solution

Emf induced e $=\mathrm{B} \ell \mathrm{v}=0.5 \times 1 \times 2=1 \mathrm{~V}$ Rate of doing work $=$ Power $\mathrm{P}=\frac{\mathrm{e}^2}{\mathrm{R}}=\frac{(1)^2}{6}=\frac{1}{6} \mathrm{~W}$ .

Asked in: MHT CET 2021 (22 Sep Shift 1)

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