JEE Advanced 2018 Paper 1 · Physics · Electromagnetic Induction and Alternating Current · medium

A straight conducting rod of mass 0.2 kg and length 0.5 m can slide without friction on two long horizontal parallel rails joined at one end by a…

One or more options correct

A straight conducting rod of mass

0.2\ \mathrm{kg}

and length

0.5\ \mathrm{m}

can slide without friction on two long horizontal parallel rails joined at one end by a resistor of

2\ \Omega

. A uniform vertical magnetic field of

0.8\ \mathrm{T}

exists everywhere. The rod, perpendicular to the rails, is given an initial speed of

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

along the rails and then left alone. Rails and rod have negligible resistance. Which of the following is/are correct?

A. The rod travels $25\ \mathrm{m}$ before coming to rest.
B. The total charge that flows through the resistor is $5\ \mathrm{C}$ .
C. The speed of the rod $2.5\ \mathrm{s}$ after the start is about $3.7\ \mathrm{m\,s^{-1}}$ .
D. By the time the speed has dropped to $5\ \mathrm{m\,s^{-1}}$ , the heat generated in the resistor is $5\ \mathrm{J}$ .

Show answer and solution

Answer: A, B, C

Motional emf gives retarding force

F=B^2L^2v/R

, so

m\,dv/dt=-kmv

with

k=\dfrac{B^2L^2}{mR}=\dfrac{0.64\times0.25}{0.2\times2}=0.4\ \mathrm{s^{-1}}

. Hence

v=v_0e^{-kt}

and the stopping distance is

v_0/k=25\ \mathrm{m}

, so (A) is true. Impulse gives

BLq=mv_0

, so

q=\dfrac{0.2\times10}{0.8\times0.5}=5\ \mathrm{C}

, so (B) is true. At

t=2.5\ \mathrm{s}

kt=1

and

v=10/e\approx3.68\ \mathrm{m\,s^{-1}}

, so (C) is true. The heat equals the loss of kinetic energy,

\tfrac12(0.2)(10^2-5^2)=7.5\ \mathrm{J}

, not

5\ \mathrm{J}

, so (D) is false.

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