MHT CET · Physics · Electromagnetic Induction
A metal conductor of length \(1 \mathrm{~m}\) rotates vertically about one of its ends at an angular velocity of \(5 \mathrm{rad} / \mathrm{s}\). if horizontal component of earth's magnetic field is \(0.2 \times 10^{-4} \mathrm{~T}\), then the e.m.f. developed between the two ends of the conductor is
- A \(5 \mu \mathrm{V}\)
- B \(50 \mathrm{mV}\)
- C \(5 \mathrm{mV}\)
- D \(50 \mu \mathrm{V}\)
Answer & Solution
Correct Answer
(D) \(50 \mu \mathrm{V}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Induced emf, } \mathrm{e}=\frac{1}{2} \mathrm{~B} \omega \ell^2 \\ & =\frac{1}{2} \times 0.2 \times 10^{-4} \times(1)^2 \times 5 \\ & =0.5 \times 10^{-4} \mathrm{~V}=50 \mu \mathrm{V}\end{aligned}\)
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