Two smooth, resistanceless rails are inclined…

Two smooth, resistanceless rails are inclined at 30^° to the horizontal and joined at the top by a 0.5 Ω resistor. A uniform magnetic field of 0.4 T…

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Two smooth, resistanceless rails are inclined at

30^\circ

to the horizontal and joined at the top by a

0.5\ \Omega

resistor. A uniform magnetic field of

0.4\ \mathrm{T}

acts vertically upward throughout the region. A rod of mass

0.1\ \mathrm{kg}

and resistance negligible,

0.5\ \mathrm{m}

long, lies across the rails perpendicular to them and slides down without friction. Taking

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

, its terminal speed is

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Answer: (C) $8.33\ \mathrm{m\,s^{-1}}$

Only the component of

B

normal to the plane,

B\cos30^\circ

, produces flux change, so

\varepsilon=Blv\cos\theta

and

I=\dfrac{Blv\cos\theta}{R}

. The magnetic force

BIl

is horizontal; its component along the incline is

BIl\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\times0.5\times0.5}{0.04\times0.75}=\dfrac{25}{3}\approx8.33\ \mathrm{m\,s^{-1}}

. If

B

were perpendicular to the incline the answer would be

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

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake