JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A wire of length \(1\,m\) moving with velocity \(8\,m / s\) at right angles to a magnetic field of \(2\,T\). The magnitude of induced emf, between the ends of wire will be \(............\,V\)
- A \(20\)
- B \(8\)
- C \(12\)
- D \(16\)
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
Induced emf across the ends \(= Bv \ell\) \(=2 \times 8 \times 1=16\,V\)
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