Question

Figure below shows a conducting rod of negligible resistance that can slide on a smooth U-shaped rail made of a wire of resistance $1\Omega m^{-1}$. Position of the conducting rod at $t=0$ is shown. A time dependent magnetic field $B=2T\text{(tesla)}$ is switched ON at $t=0$.
Following the situation of the previous question, the magnitude of force required to move the conducting rod at a constant speed of $5{\text{cms}}^{-1}$ at the same instant $t=2s$ is equal to

• A. 0.16 N
• B. 0.12 N
• C. 0.08 N
• D. 0.06 N

Answer 1

The correct option is C 0.08 N

When a current carrying loop is placed in magnetic field it experiences a force given by,
$F=I(\overrightarrow{l}\times \overrightarrow{B})$
In terms of magnitude,
$\begin{array}{}F=BIl& \text{(1)}\end{array}$
Current through the loop after 2 sec when the rod is moving with velocity 5 cm/s,
$I=\frac{E}{R}$
From Faraday's law induced emf, $E=-\frac{d\varphi }{dt}$
Since flux, $\varphi =B.A$
After $t$ sec if wire is moving with velocity $v$ area, $A=0.2\times (0.4-v\times t)$
Also magnetic field, $B=2t$
Therefore flux, $\varphi =B.A=2t\times 0.2\times (0.4-v\times t)$
$\Rightarrow \varphi =0.16t-0.4v{t}^2$
$\frac{d\varphi }{dt}=0.16-0.8vt$
At $t=2$ sec and $v=5$ cm/s = $0.05$ m/s
$\frac{d\varphi }{dt}=0.16-0.8\times 0.05\times 2=0.08$
$\Rightarrow E=-\frac{d\varphi }{dt}=-0.08V$
In terms of magnitude, $E=0.08V$
Length of the loop after 2 sec, $L=0.2+2\times (0.4-v\times t)$
$L=0.2+2\times (0.4-0.05\times 2)=0.8m$
Therefore resistance, $R=1\Omega /m\times 0.8=0.8\Omega$
Thus current, $I=\frac{E}{R}=\frac{0.08}{0.8}=0.1A$
Length of the rod, $l=0.2m$
Since magnetic field, $B=2t$
Afetr 2 sec magnetic field, $B=2\times 2=4T$
Substituting the values of $B,I$ and $l$ in $\left(1\right)$,
Force required to move conducting rod, $F=BIl=4\times 0.1\times 0.2=0.08N$