JEE Advanced 2023 Paper 2 · Physics · Electromagnetic Induction and Alternating Current · hard

A thin uniform metal disc of radius a=0.50 m can rotate without friction about its geometric axis, which is parallel to a uniform magnetic field…

Numerical answer

A thin uniform metal disc of radius

a=0.50\ \mathrm{m}

can rotate without friction about its geometric axis, which is parallel to a uniform magnetic field

B=2.0\ \mathrm{T}

perpendicular to the plane of the disc. A resistor

R=1.0\ \Omega

is connected between the axle and the rim through sliding contacts (the resistances of the disc and the contacts are negligible). A light string wound on the axle, of radius

r=0.10\ \mathrm{m}

, carries a mass

m=0.50\ \mathrm{kg}

that descends and finally reaches a constant speed. Take

g=10\ \mathrm{m\,s^{-2}}

. Find the final angular speed of the disc in

\mathrm{rad\,s^{-1}}

Show answer and solution

Answer: 8

When the disc turns at

\omega

, each radial element sees

v=\omega x

, so the emf between axle and rim is

\varepsilon=\int_0^aB\omega x\,dx=\tfrac12B\omega a^2

and the current is

i=\varepsilon/R

. The magnetic force on the radial current has torque

\tau_B=\int_0^aiBx\,dx=\tfrac12iBa^2=\dfrac{B^2a^4\omega}{4R}

, opposing the rotation. At constant speed this balances the torque of the hanging mass,

mgr

. So

\omega=\dfrac{4mgrR}{B^2a^4}=\dfrac{4\times0.5\times10\times0.1\times1}{4\times0.0625}=\dfrac{2}{0.25}=8\ \mathrm{rad\,s^{-1}}

. As a check,

i=\tfrac12\times2\times8\times0.25=2\ \mathrm{A}

and

i^2R=4\ \mathrm{W}=mg\,(\omega r)=5\times0.8

, so the heat generated equals the power of gravity.

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