A rod of length ' $l$ ' is rotated with angular velocity ' $\omega$ ' about its one end, perpendicular to a magnetic field of induction 'B'. The e.m.f. induced in the rod is

• A

  $\mathrm{B} l^2 \omega$
• Correct answer
  B

  $0.5 \mathrm{~B} l^2 \omega$
• C

  $\mathrm{B} / \omega$
• D

  $0.5 \mathrm{~B} / \omega$

Solution

A conducting rod of length ' $l$ ' whose one end is fixed, is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocity $\omega$.
E.M.F induced, $\mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n}=\frac{\mathrm{B} \pi l^2}{\mathrm{~T}}$ $\therefore \quad \mathrm{e}=\frac{1}{2} \mathrm{~B} l^2 \omega=0.5 \mathrm{~B} l^2 \omega \quad \ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}\right)$

Hence, the correct answer is option: 'B'.