JEE Advanced 2024 Paper 2 · Physics · Magnetic Effects of Current · medium

Two long, parallel, smooth horizontal conducting rails, a distance l=2 m apart, are connected at one end by an uncharged capacitor of capacitance…

Single correct option

Two long, parallel, smooth horizontal conducting rails, a distance

l=2\ \mathrm{m}

apart, are connected at one end by an uncharged capacitor of capacitance

C=0.3\ \mathrm{F}

. A uniform magnetic field

B=0.5\ \mathrm{T}

acts vertically downward. A conducting rod of mass

m=0.2\ \mathrm{kg}

is placed across the rails at rest and is pulled by a constant horizontal force

F=1.5\ \mathrm{N}

perpendicular to its length. The resistance of rod and rails is negligible. The charge on the capacitor at

t=2\ \mathrm{s}

Show answer and solution

Answer: (C) $1.8\ \mathrm{C}$

With no resistance, the capacitor voltage equals the motional emf:

q=CBlv

. The current is

i=dq/dt=CBl\,a

, giving a retarding magnetic force

Bil=CB^2l^2a

on the rod. Newton's law:

F-CB^2l^2a=ma

, so

a=\dfrac{F}{m+CB^2l^2}

. Here

Bl=1\ \mathrm{T\,m}

and

CB^2l^2=0.3\ \mathrm{kg}

, so

a=\dfrac{1.5}{0.5}=3\ \mathrm{m\,s^{-2}}

(constant). At

t=2\ \mathrm{s}

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

and

q=CBlv=0.3\times1\times6=1.8\ \mathrm{C}

. Option (A) neglects the back-force (uses

a=F/m

), (B) drops the factor

B

(uses

Clv

), and (D) uses the average rather than final velocity. The correct option is (C).

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake