JEE Main 2025 Jan 28 Shift 2, Physics Q1: Motional EMF

JEE Main2025Jan 28, Shift 2Physics

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 protential difference developed between the axis of the disc and the rim ? ( $π = 3.14$ )

A
$0.0628$ V
B
$0.5024$ V
C
$0.2512$ V
D
$0.1256$ V

Solution

For a conducting disc rotating in a perpendicular magnetic field, the motional EMF between the axis and the rim is

$ε = \frac{1}{2} B ω R^{2}$ .

Substituting $B = 0.4$ T, $ω = 10 π$ rad/s and $R = 0.20$ m:

$ε = \frac{1}{2} \times 0.4 \times 10 π \times (0.2)^{2} = 0.08 π \approx 0.2512$ V.

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JEE Main 2025 Jan 28 Shift 2, Physics Q1: Motional EMF

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