JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2023
- Medium
- MCQ
A conducting loop of radius \(\frac{10}{\sqrt{\pi}}\,cm\) is placed perpendicular to a uniform magnetic field of \(0.5\,T\). The magnetic field is decreased to \(zero\) in \(0.5\,s\) at a steady rate. The induced emf in the circular loop at \(0.25\,s\) is:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(emf =10\,mV\)
Step-by-step Solution
Detailed explanation
\(EMF =\frac{ d \phi}{ dt }=\frac{ BA -0}{ t }\)
\(A =\pi r ^2=\pi\left(\frac{0.1^2}{\pi}\right)=0.01\)
\(B =0.5\)
\(EMF =\frac{(0.5)(0.01)}{0.5}=0.01\,V =10\,mV\)
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