MHT CET · Physics · Electromagnetic Induction
A metal rod of length ' \(l\) ' rotates about one of its ends in a plane perpendicular to a magnetic field of induction ' B '. If the e.m.f. induced between the ends of the rod is ' \(e\) ', then the number of revolutions made by the rod per second is
- A \(\frac{\mathrm{e}}{\mathrm{B} \pi^2 l}\)
- B \(\frac{\mathrm{e}}{\mathrm{B} \pi l^2}\)
- C \(\frac{\mathrm{B}^2}{\mathrm{e} \pi l}\)
- D \(\frac{\pi l^2}{\mathrm{eB}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{e}}{\mathrm{B} \pi l^2}\)
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 \(\omega\).
\(\begin{aligned} & \mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n} \\ \therefore \quad \mathrm{n} & =\frac{\mathrm{e}}{\pi \mathrm{B} l^2}\end{aligned}\)
\(\begin{aligned} & \mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n} \\ \therefore \quad \mathrm{n} & =\frac{\mathrm{e}}{\pi \mathrm{B} l^2}\end{aligned}\)
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