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MHT CET · Physics · Electromagnetic Induction

When a rod of length \(l\) is rotated with angular velocity of \(\omega\) in a perpendicular field of induction \(B\), about one end, the emf across its ends is

  1. A \(B l^{2} \omega\)
  2. B \(\frac{B l^{2} \omega}{2}\)
  3. C \(Bl\omega\)
  4. D \(\frac{B l \omega}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{B l^{2} \omega}{2}\)

Step-by-step Solution

Detailed explanation

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 \(\omega\). Magnetic field \((B)\) is perpendicular to the plane of the paper. Emf induced across the ends of the rod is \(e=B A n\)

\(
\begin{array}{l}
=B \pi l^{2} n \\
=\frac{B l^{2} \pi}{T} \\
=\frac{1}{2} B l^{2} \omega
\end{array}
\)
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