Conductor wire ABCDE with each arm 10 cm in length is placed in… | JEE Main 2025 (4 Apr, Shift 1)

Q14JEE Main 2025 · 4 APR · SHIFT 1Medium

Motional EMF

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 $10\text{ cm/s}$ , induced emf between points A and E is ________ mV.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

Emf between A and E equals that of the straight line AE, using its length perpendicular to $\vec v$ .
AB and DE are along $\vec v$ ; BC and CD are at $45^\circ$ , each giving $10\sin45^\circ = \frac{10}{\sqrt2}$ cm across.
Effective length $= 2\times\frac{10}{\sqrt2} = 10\sqrt2$ cm $= 0.1\sqrt2$ m.
$\varepsilon = \frac{1}{\sqrt2}\times0.1\sqrt2\times0.1 = 0.01$ V
Answer: $10$ mV.

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