Question

An electric circuit is composed of the three conducting rods $\text{MO},\text{ON}$ and $\text{PQ}$ as shown in the figure. The resistance of the rods per unit length is $\lambda$. The rod $\text{PQ}$ slides, as shown in the figure, at a constant velocity $v$, keeping its tilt angle relative to $\text{ON}$ and $\text{MO}$ fixed at ${45}^{\circ }$. At each instance, the circuit is closed. The whole system is embedded in a uniform magnetic field $B$, which is directed perpendicularly into the page. Compute the time dependent induced electric current.

• A. $\frac{Bv}{\lambda (1+\sqrt{2})}$
• B. $\frac{2Bv}{\lambda (1+\sqrt{2})}$
• C. $\frac{Bv}{\lambda }$
• D. Zero

Answer 1

The correct option is A $\frac{Bv}{\lambda (1+\sqrt{2})}$

The position of the moving rod at any time $t$ is shown in the figure:


From the figure, we can say that,

$OR=vt,$ distance travelled by rod in time $t$.

$O{Q}^{\prime }\cos {45}^{\circ }=OR$

$\Rightarrow O{Q}^{\prime }=OR\sqrt{2}=vt\sqrt{2}$

And $P^{\prime }O=OQ=vt\sqrt{2}$

$\therefore$ The length of rod $P^{\prime }{Q}^{\prime }$ which is in contact with others rod at any time $t$ is,

$P^{\prime }{Q}^{\prime }=\sqrt{O{P}^{\prime 2}+O{Q}^{\prime 2}}=\sqrt{(\sqrt{2}vt{)}^2+(\sqrt{2}vt{)}^2}=2vt$

$\therefore$ Emf induced across $P^{\prime }{Q}^{\prime }$ is,

$E=B\times \left(2vt\right)\times v=2B{v}^{2}t$

Total resistance of the loop is,

$R_{total}=\lambda (O{P}^{\prime }+O{Q}^{\prime }+P{Q}^{\prime })$

$=\lambda (2vt+\sqrt{2}vt+\sqrt{2}vt)$

$=2\lambda vt(1+\sqrt{2})$

$\therefore$ Induced correct in circuit is,

$i=\frac{E}{R_{total}}=\frac{2B{v}^{2}t}{2\lambda vt(1+\sqrt{2})}=\frac{Bv}{\lambda (1+\sqrt{2})}$

Hence, $\left(\text{A}\right)$ is the correct answer.