A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of…

A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \mathrm{~T}$ with a frequency of $12 \mathrm{rev} / \mathrm{s}$. The induced emf across the ends to the rod is
  1. $18.89 \mathrm{~V}$
  2. $3 \mathrm{~V}$
  3. $15 \mathrm{~V}$
  4. $9.42 \mathrm{~V}$

Solution

Length of rod, $l=1.0 \mathrm{~m}$ $ \begin{aligned} B=0.25 \mathrm{~T}, f & =12 \mathrm{rev} / \mathrm{s} \\ \therefore \quad \text { Induced emf, } e & =\frac{B \omega l^2}{2}=\frac{0.25 \times 2 \pi f}{2} \times 1^2 \\ & =0.25 \pi f=0.25 \pi \times 12 \\ & =3 \pi=9.42 \mathrm{~V} \end{aligned} $

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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