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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2025 · 28 January · Evening Shift · Physics · Electromagnetic Induction

A uniform magnetic field of 0.4 T acts perpendicular to a circular...

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 10πrads−110 \pi \mathrm{rad} \mathrm{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)(\pi=3.14)

Options

  1. A0.5024 V0.5024 \mathrm{~V}
  2. B V\mathrm{~V}
  3. C0.2512 V0.2512 \mathrm{~V}
  4. D0.1256 V0.1256 \mathrm{~V}

Correct answer

C. 0.2512 V0.2512 \mathrm{~V}

Step-by-step solution

B=0.4 Tr=20 cmω=10πrad/sE=12 BωR2=0.2512 V\begin{aligned} & \mathrm{B}=0.4 \mathrm{~T} \\ & \mathrm{r}=20 \mathrm{~cm} \\ & \omega=10 \pi \mathrm{rad} / \mathrm{s} \\ & \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^2 \\ & =0.2512 \mathrm{~V}\end{aligned}
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