JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2019
- Hard
- Numerical
A conducting circular loop made of a thin wire, has area \(3.5 \times 10^{-3}\, m^2\) and resistance \(10\,\Omega \). It is placed perpendicular to a time dependent magnetic field \(B(t) = (0.4\,T)\, sin\, (50\, \pi t)\). The field is uniform in space. Then the net charge flowing through the loop during \(t = 0\, s\) and \(t = 10\, ms\) is close to.......\(mC\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(140\)
Step-by-step Solution
Detailed explanation
\({\text{B}}({\text{t}}) = 0.4\sin \left( {50\pi \times {{10}^{ - 2}}} \right)\)
\( = 0.4\sin \left( {\frac{{50\pi }}{{100}}} \right) = 0.4\)
\(\Delta q = \frac{{ - \Delta {\text{Q}}}}{{\text{R}}} = \frac{{\Delta {\text{Q}}}}{{\text{R}}}\)
\( = \frac{{0.4 \times 3.5 \times {{10}^{ - 3}}}}{{10}}\)
\( = 140\,{\text{mc}}\)
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