Question

A long conducting wire $AH$ is moved over a conducting triangular wire $CDE$ with a constant velocity $v$ in a uniform magnetic field $\overrightarrow{B}$ directed into the plane of the paper. Resistance per unit length of each wire $\rho$. Then

• A. a constant clockwise induced current will flow in the closed loop
• B. an increasing anticlockwise induced current will flow in the closed loop
• C. a decreasing anticlockwise induced current will flow in the closed loop
• D. a constant anticlockwise induced current will flow in the closed loop

Answer 1

The correct option is D a constant anticlockwise induced current will flow in the closed loop

Intial flux = $\frac{1}{2}xhB$

flux at time t = $\frac{1}{2}\left(\frac{x}{h}\right(h+vt\left)\right)(h+vt)B=\frac{1}{2}\frac{x}{h}(h+vt{)}^{2}B$

emf=rate of change of flux= $\frac{x}{h}(h+vt)v$

length of loop at time t = $\frac{x}{h}(h+vt)+2\frac{y}{h}(h+vt)=k(h+vt)$ ;$k=\frac{x}{h}+\frac{2y}{h}$

resistance of loop = $\rho k(h+vt)$

current at time t = $\frac{emf}{resistance}$= independent of t.

As the flux keeps increases through the flux , by Lenz's Law, induced current will be in the anticlockwise direction. And from the above calculations, current will be constant.