Question

A conducting rod of mass m and length l is given a velocity $V_0$ along the rails as shown is the figure. Magnetic field B exists perpendicular to the plane, then [Consider rails to be very long]

• A. Time taken by rod to come to rest is $\frac{mR}{l^2{B}^2}$.
• B. Distance travelled by rod before stopping is $\frac{m{v}_{0}R}{l^2{B}^2}$.
• C. Heat generated across R till rod stops is $\frac{m{v}_0^2}{2}$.
• D. Distance travelled by rod before stopping is infinite.

Answer 1

Answer: (B), (C)

The correct options are
B Distance travelled by rod before stopping is $\frac{m{v}_{0}R}{l^2{B}^2}$.
C Heat generated across R till rod stops is $\frac{m{v}_0^2}{2}$.


$i=\frac{\epsilon }{R}=\frac{Bvl}{R}$
$\Rightarrow F=-ilB$ (Deceleration)
$a=\frac{F}{m}=-\frac{ilB}{m}$
$a=-\frac{l^2{B}^{2}v}{mR}\Rightarrow {\int }_{v_0}^v\frac{dv}{v}$
= -${\int }_0^t\frac{l^2{B}^2}{mR}dt$
$ln\frac{v}{v_0}=\frac{-l^2{B}^{2}t}{mR}$
$\Rightarrow v=v_{0}e\frac{-l^2{B}^{2}t}{mR}$ ...(1)
For $v=0,t=\infty$.
$a=\frac{vdv}{dx}\Rightarrow \frac{\text{/}vdv}{dx}$ =- $\frac{l^2{B}^2\text{/}v}{mR}$
${\int }_{v_0}^{v}dv=\frac{-l^2{B}^2}{mR}{\int }_0^{x}dx$ \
$\Rightarrow v=v_0-\frac{l^2{B}^2}{mR}x$
$v=0\Rightarrow x=\frac{mR{v}_0}{B^2{l}^2}$
The initial KE of the rod is $\frac{1}{2}m{v}_0^2$. This will be the heat lost after the rod comes to rest.