JEE Main 2021 · Physics · Magnetic Effects of Current · hard

Two long smooth conducting rails are inclined at 30^° to the horizontal and are 0.5 m apart. They are joined at the lower end by a 3 Ω resistor. A…

Numerical answer

Two long smooth conducting rails are inclined at

30^\circ

to the horizontal and are

0.5\ \mathrm{m}

apart. They are joined at the lower end by a

3\ \Omega

resistor. A uniform field of

2\ \mathrm{T}

is perpendicular to the plane of the rails. A rod of mass

0.4\ \mathrm{kg}

slides down the rails remaining perpendicular to them. Neglecting other resistance and taking

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

, the power dissipated in the resistor at terminal speed, in watts, is

Show answer and solution

Answer: 12

At terminal speed

mg\sin\theta=BIL=\dfrac{B^2L^2v}{R}

. With

mg\sin\theta=0.4\times10\times0.5=2\ \mathrm{N}

and

B^2L^2=4\times0.25=1

v=\dfrac{2\times3}{1}=6\ \mathrm{m\,s^{-1}}

. Power dissipated equals the rate of loss of gravitational PE:

P=mg\sin\theta\,v=2\times6=12\ \mathrm{W}

Check:

BLv=2\times0.5\times6=6\ \mathrm{V}

and

V^2/R=36/3=12\ \mathrm{W}

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake