MHT CET · Physics · Electromagnetic Induction
A wire of length ' \(L\) ' having resistance ' \(R\) ' falls from a height ' \(\ell\) ' in earth's horizontal magnetic field ' \(\mathrm{B}\) '. The induced emf through the wire is ( \(\mathrm{g}\) = acceleration due to gravity)
- A \(\mathrm{BL} \sqrt{2 \mathrm{~g} \ell}\)
- B \(\frac{\mathrm{BL} \sqrt{2 \mathrm{~g} \ell}}{2}\)
- C \(\frac{\mathrm{BL} \sqrt{2 \mathrm{~g} \ell}}{\mathrm{R}}\)
- D \(\frac{\mathrm{BL}}{\sqrt{2 \mathrm{~g} \ell}}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{BL} \sqrt{2 \mathrm{~g} \ell}\)
Step-by-step Solution
Detailed explanation
If the wire falls through a height \(\ell\), the velocity acquired by it is
\(
\mathrm{v}=\sqrt{2 \mathrm{~g} \ell}
\)
The emf induced in the wire
\(
\mathrm{E}=\mathrm{BLv}=\mathrm{BL} \sqrt{2 \mathrm{~g} \ell}
\)
\(
\mathrm{v}=\sqrt{2 \mathrm{~g} \ell}
\)
The emf induced in the wire
\(
\mathrm{E}=\mathrm{BLv}=\mathrm{BL} \sqrt{2 \mathrm{~g} \ell}
\)
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