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 $10 \pi \mathrm{rad} \mathrm{s}^{-1}$ 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 ? ( $\pi=3.14$ )

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Electromagnetic Induction

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✓ Correct
(D)0.1256 V

Answer: (C)

Solution

A conducting disc of radius

R

rotating with angular velocity

ω

in a uniform field

B

perpendicular to its plane develops an EMF between its centre and rim equal to

E = \frac{1}{2} B ω R^{2}

(this follows from summing the motional EMF

B ω r d r

of each elementary ring from the centre to the rim).

Substituting

B = 0.4 T

ω = 10 π rad/s = 31.4 rad/s

R = 0.2 m

E = \frac{1}{2} \times 0.4 \times 31.4 \times (0.2)^{2} = \frac{1}{2} \times 0.4 \times 31.4 \times 0.04

E = 0.2512 V

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Electromagnetic Induction

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