Question

Conductor wire ABCDE with each arm 10 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.}

Hint:

Use Faraday's law to calculate the induced emf.

Answer 1

Correct Answer: 10

1. Induced emf: \[ \varepsilon = Bv \ell_{\mathrm{AB}} \] \[ \varepsilon = \frac{1}{\sqrt{2}} \times \frac{10 \mathrm{~cm}}{s} \times 2 \left( 10 \sin 45^{\circ} \right) \mathrm{cm} \] \[ \varepsilon = 10 \mathrm{~mV} \] Therefore, the correct answer is (10) 10 mV.