JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A rectangular loop of length \(2.5 \mathrm{~m}\) and width \(2 \mathrm{~m}\) is placed at \(60^{\circ}\) to a magnetic field of \(4 \mathrm{~T}\). The loop is removed from the field in \(10 \ \mathrm{sec}\). The average emf induced in the loop during this time is
- A \(-2 \mathrm{~V}\)
- B \(+2 \mathrm{~V}\)
- C \(+1\mathrm{ V}\)
- D \(-1 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(+1\mathrm{ V}\)
Step-by-step Solution
Detailed explanation
\( \text { Average emf }=\frac{\text { Change in flux }}{\text { Time }}=-\frac{\Delta \phi}{\Delta t} \) \( =-\frac{0-\left(4 \times(2.5 \times 2) \cos 60^{\circ}\right)}{10} \) \( =+1 \mathrm{~V}\)
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