A uniform metal rod of length l=0.5 m…

A uniform metal rod of length l=0.5 m rotates about one end with constant angular velocity ω=40 rad s^-1 in a uniform magnetic field B=0.2 T…

One or more options correct

A uniform metal rod of length

l=0.5\ \mathrm m

rotates about one end with constant angular velocity

\omega=40\ \mathrm{rad\,s^{-1}}

in a uniform magnetic field

B=0.2\ \mathrm T

perpendicular to the plane of rotation. The free end slides on a stationary conducting ring of radius

l

centred at the pivot, and a resistor

R=0.2\ \Omega

connects the pivot to the ring (all other resistances and friction are negligible). Which of the following is/are correct?

A. The emf induced in the rod is $2.0\ \mathrm V$ .
B. The external torque needed to maintain the rotation is $0.25\ \mathrm{N\,m}$ .
C. The mechanical power supplied by the external agent is $5\ \mathrm W$ .
D. If both $B$ and $\omega$ are doubled, the external torque needed is doubled.

Show answer and solution

Answer: C

The emf is

\mathcal E=\tfrac12B\omega l^2=\tfrac12(0.2)(40)(0.25)=1.0\ \mathrm V

, so (A) is incorrect (that value forgets the factor

\tfrac12

). Current

I=\mathcal E/R=5\ \mathrm A

. The magnetic force

BIl

on the rod acts at its midpoint, giving a retarding torque

\tau=BIl\cdot\dfrac l2=\tfrac12BIl^2=\tfrac12(0.2)(5)(0.25)=0.125\ \mathrm{N\,m}

; (B) is incorrect. Power supplied

P=\tau\omega=0.125\times40=5\ \mathrm W

, equal to the Joule heating

I^2R=25\times0.2=5\ \mathrm W

: (C) is correct. In general

\tau=\dfrac{B^2l^4\omega}{4R}

(since

I=\dfrac{B\omega l^2}{2R}

), so doubling both

B

and

\omega

multiplies the torque by

2^2\times2=8

, not

2

: (D) is incorrect.

Original question written in the JEE Advanced 2022 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake