MHT CET · Physics · Electromagnetic Induction
A straight conductor of length \(0.6 \mathrm{~m}\) is moved with a speed of \(10 \mathrm{~ms}^{-1}\) perpendicular to magnetic field of induction 1.2 weber \(\mathrm{m}^{-2}\). The induced e.m.f. across the conductor is
- A \(6 \mathrm{~V}\)
- B \(7.2 \mathrm{~V}\)
- C \(0.72 \mathrm{~V}\)
- D \(12 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(7.2 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The induced emf is given by
\(
\mathrm{E}=\mathrm{B} \ell \mathrm{v}=1.2 \times 0.6 \times 10=7.2 \mathrm{~V}
\)
\(
\mathrm{E}=\mathrm{B} \ell \mathrm{v}=1.2 \times 0.6 \times 10=7.2 \mathrm{~V}
\)
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