MHT CET · Physics · Electromagnetic Induction
A wire of length \(1 \mathrm{~m}\) is moving at a speed of \(2 \mathrm{~m} / \mathrm{s}\) perpendicular homogenous magnetic field of \(0.5 \mathrm{~T}\). The ends of the wire are joined to resistance \(6 \Omega\). The rate at which work is being done to keep the wire moving at that speed is
- A \(\frac{1}{3} \mathrm{~W}\)
- B \(\frac{1}{6} \mathrm{~W}\)
- C \(\frac{1}{12} \mathrm{~W}\)
- D \(1 \mathrm{~W}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{6} \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
Emf induced e \(=\mathrm{B} \ell \mathrm{v}=0.5 \times 1 \times 2=1 \mathrm{~V}\)
Rate of doing work \(=\) Power
\(
\mathrm{P}=\frac{\mathrm{e}^2}{\mathrm{R}}=\frac{(1)^2}{6}=\frac{1}{6} \mathrm{~W}
\)
Rate of doing work \(=\) Power
\(
\mathrm{P}=\frac{\mathrm{e}^2}{\mathrm{R}}=\frac{(1)^2}{6}=\frac{1}{6} \mathrm{~W}
\)
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