MHT CET · Physics · Electromagnetic Induction
A magnetic field of \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(100 \mathrm{~cm}^2\) with 50 turns. The average e.m.f. induced in the coil is \(0.1 \mathrm{~V}\), when it is removed from the field in time ' \(t\) '. The value of ' \(t\) ' is
- A \(2 \times 10^{-3} \mathrm{~s}\)
- B \(0.5 \mathrm{~s}\)
- C \(0.1 \mathrm{~s}\)
- D \(1 \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(0.1 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\left(\phi_2-\phi_1\right)}{\mathrm{t}}=-\frac{(0-\mathrm{NBA})}{\mathrm{t}}\)
\(\therefore 0.1=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{\mathrm{t}}\)
\(\therefore\) The value of ' \(t\) ' is,
\(\mathrm{t}=\frac{10^{-2}}{0.1}=0.1 \mathrm{~s}\)
\(\therefore 0.1=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{\mathrm{t}}\)
\(\therefore\) The value of ' \(t\) ' is,
\(\mathrm{t}=\frac{10^{-2}}{0.1}=0.1 \mathrm{~s}\)
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