MHT CET · Physics · Electromagnetic Induction
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 \(\mathrm{B} l^2 \omega\)
- B \(0.5 \mathrm{~B} l^2 \omega\)
- C \(\mathrm{B} / \omega\)
- D \(0.5 \mathrm{~B} / \omega\)
Answer & Solution
Correct Answer
(B) \(0.5 \mathrm{~B} l^2 \omega\)
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 \(\omega\).
E.M.F induced, \(\mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n}=\frac{\mathrm{B} \pi l^2}{\mathrm{~T}}\)
\(\therefore \quad \mathrm{e}=\frac{1}{2} \mathrm{~B} l^2 \omega=0.5 \mathrm{~B} l^2 \omega \quad \ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}\right)\)
E.M.F induced, \(\mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n}=\frac{\mathrm{B} \pi l^2}{\mathrm{~T}}\)
\(\therefore \quad \mathrm{e}=\frac{1}{2} \mathrm{~B} l^2 \omega=0.5 \mathrm{~B} l^2 \omega \quad \ldots\left(\because \omega=\frac{2 \pi}{\mathrm{~T}}\right)\)
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