JEE Advanced 2022 Paper 1 · Physics · Magnetic Effects of Current · hard

Two parallel conducting rails, 0.5 m apart, are inclined at 30^° to the horizontal and joined at the lower end by a resistor of 0.2 Ω. A smooth…

Numerical answer

Two parallel conducting rails,

0.5\ \mathrm{m}

apart, are inclined at

30^\circ

to the horizontal and joined at the lower end by a resistor of

0.2\ \Omega

. A smooth conducting rod of mass

0.1\ \mathrm{kg}

lies across them, perpendicular to the rails. A uniform magnetic field of

0.4\ \mathrm{T}

acts vertically upward everywhere. The rod is released from rest and the rails are long. Take

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

and neglect the resistance of rails and rod. The terminal speed of the rod in

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

is ______ . Give the answer correct to two decimal places.

Show answer and solution

Answer: 3.33

The rod moves along the incline with speed

v

. Only the component of

\vec B

perpendicular to the incline,

B\cos30^\circ

, is effective, so the emf is

\varepsilon=B l v\cos\theta

and the current is

I=Blv\cos\theta/R

. The magnetic force

IlB

is horizontal; its component along the incline is

IlB\cos\theta

. At terminal speed

mg\sin\theta=\dfrac{B^2l^2v\cos^2\theta}{R}

. So

v=\dfrac{mgR\sin\theta}{B^2l^2\cos^2\theta}=\dfrac{0.1\times10\times0.2\times0.5}{0.16\times0.25\times0.75}=\dfrac{0.1}{0.03}=3.33\ \mathrm{m\,s^{-1}}

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