Two long, smooth, horizontal parallel…

Two long, smooth, horizontal parallel rails are 0.5 m apart in a uniform vertical magnetic field B=0.8 T. They are joined at one end by an ideal…

One or more options correct

Two long, smooth, horizontal parallel rails are

0.5\ \mathrm{m}

apart in a uniform vertical magnetic field

B=0.8\ \mathrm{T}

. They are joined at one end by an ideal capacitor of capacitance

1\ \mathrm{F}

, initially uncharged. A uniform rod of mass

0.2\ \mathrm{kg}

lies across the rails and is pulled from rest by a constant horizontal force

F=1\ \mathrm{N}

along the rails. The rails and rod have negligible resistance. Which of the following statements is/are correct?

A. The acceleration of the rod is constant and equal to about $2.78\ \mathrm{m\,s^{-2}}$ .
B. The current in the circuit is constant and equal to about $1.11\ \mathrm{A}$ .
C. At $t=3\ \mathrm{s}$ the energy stored in the capacitor is about $6.94\ \mathrm{J}$ .
D. At $t=3\ \mathrm{s}$ all the work done by the force $F$ up to that time is stored as kinetic energy of the rod.

Show answer and solution

Answer: A, B

The capacitor voltage equals the emf,

q=CBLv

, so the current is

i=CBL\,a

. Newton's law:

F-iBL=ma

gives

a=\dfrac{F}{m+CB^2L^2}=\dfrac{1}{0.2+0.16}=2.78\ \mathrm{m\,s^{-2}}

, which is constant. Hence

i=CBLa=1\times0.8\times0.5\times2.78=1.11\ \mathrm{A}

, constant. A and B are correct. At

t=3\ \mathrm{s}

v=at=8.33\ \mathrm{m\,s^{-1}}

and the emf is

BLv=3.33\ \mathrm{V}

, so the capacitor energy is

\tfrac12C(BLv)^2=5.56\ \mathrm{J}

. The value

6.94\ \mathrm{J}

is the kinetic energy

\tfrac12mv^2

of the rod, so C is wrong. The work done by

F

F\cdot\tfrac12at^2=12.5\ \mathrm{J}=5.56+6.94

, so it is shared between the capacitor and the rod; D is wrong.

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake