JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2021
- Medium
- Numerical
A circular coil of radius \(8.0 \,{cm}\) and \(20\) turns is rotated about its vertical diameter with an angular speed of \(50\, {rad} {s}^{-1}\) in a uniform horizontal magnetic field of \(3.0 \times 10^{-2}\, {T}\). The maximum \(emf\) induced the coil will be \(\ldots \ldots \ldots \times 10^{-2}\) \(volt\) (rounded off to the nearest integer)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
Maximum emf \(\varepsilon={N} \omega {AB}\)
\({N}=20, \omega=50, {B}=3 \times 10^{-2} {T}\)
\(\varepsilon=20 \times 50 \times \pi \times(0.08)^{2} \times 3 \times 10^{-2}=60.28 \times 10^{-2}\)
Rounded off to nearest integer \(=60\)
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