Q78 / 90 · KVPY 2016 (SB/SX) · Physics
A conducting bar of mass $\mathrm{m}$ and length $\ell$ moves on two frictionless parallel rails in the presence of a constant uniform magnetic field of magnitude B directed into the page as shown in the figure . The bar is given an initial velocity $\mathrm{v}_{0}$ towards the right at $\mathrm{t}=0$. Then the
- AInduced current in the circuit is in the clockwise direction
- BVelocity of the bar decreases linearly with time
- Correct answerCDistance the bar travels before it comes to a complete stop is proportional to $\mathrm{R}$
- DPower generated across the resistance is proportional to $\ell$.
Solution
$\mathrm{F}=\mathrm{iB} \ell$
$\mathrm{a}=\frac{\mathrm{iB} \ell}{\mathrm{m}}$
$\phi=\mathrm{B} . \mathrm{A}$
$\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{B} \cdot \ell \cdot\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)$
$\varepsilon=(\mathrm{B} \cdot \ell \mathrm{v})$
$\mathrm{i}=\varepsilon / \mathrm{R}=\frac{\mathrm{B} \cdot \ell \mathrm{v}}{\mathrm{R}}$
$\mathrm{a}=\left(\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}\right) \frac{\mathrm{B} \ell}{\mathrm{m}}$
$\mathrm{a}=\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{Rm}} \cdot \mathrm{v}$
$\Rightarrow \mathrm{a}=\mathrm{v} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$
$\Rightarrow \mathrm{v} . \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{Rm}} \cdot \mathrm{v}$
$\Rightarrow \int \mathrm{dv}=\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{Rm}} \cdot \int \mathrm{d} \mathrm{x}$
$\mathrm{v}=\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{Rm}} \cdot{ }^{\prime} \mathrm{X}^{\prime}\left[\mathrm{X}=\frac{\mathrm{vRM}}{\mathrm{B}^{\prime} \ell^{2}}\right]$
Hence, the correct answer is option: 'C'.