JEE Mains · Physics · STD 12 -6. Electromagnetic induction
Conductor wire ABCDE with each \(\operatorname{arm} 10 \mathrm{~cm}\) in length is placed in magnetic field of \(\frac{1}{\sqrt{2}}\) Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of \(10 \mathrm{~cm} / \mathrm{s}\), induced emf between points A and E is _____ mV.
- A 10
- B 15
- C 20
- D 25
Answer & Solution
Correct Answer
(A) 10
Step-by-step Solution
Detailed explanation
As field is uniform we can replace the bent wire with straight wire from \(A\) to \(B\). So EMF : \(\varepsilon=\mathrm{Bv} \ell_{\mathrm{AB}}\)…
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