A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 π rad s-1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π = 3.14) (a) 0.5024 V (b) 0.0628 V (c) 0.2512 V (d) 0.1256 V

Answer: (c). e = 1/2Bω R² e = 1/2 × 0.4 × 10π × (0.2)² = 0.08π e = 0.08 × 3.14 = 0.2512 V, so the answer is (c).

A uniform magnetic field of 0.4 T acts perpendicular to a circular… | JEE Main 2025 (28 Jan, Shift 2)

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Rotating Conductors & AC Generator

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 $\pi$ rad s-1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? $(\pi = 3.14)$

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Answer: (c)

$e = \tfrac{1}{2}B\omega R^2$
$e = \tfrac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2 = 0.08\pi$
$e = 0.08 \times 3.14 = 0.2512\text{ V}$ , so the answer is (c).

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