Question

A rectangular loop with a sliding conductor of length $l$ is located in a uniform magnetic field perpendicular to the plane of the loop (figure). The magnetic induction is $B$. The conductor has a resistance $R$. The sides $A B$ and $C D$ have resistances $R_{1}$ and $R_{2}$, respectively. Find the current through the conductor during its motion to the right with a constant velocity $v$.

• A. $\frac{B l v\left(R_{1}+R_{2}\right)}{R_{1}\left(R_{1}+R_{2}\right)}$
• B. $\frac{B l^{2} v}{R_{1}+R_{1} R_{2}}$
• C. $\frac{B l v\left(R_{1}+R_{2}\right)}{R_{1} R_{2}+R\left(R_{1}+R_{2}\right)}$
• D. $\frac{B l^{2} v}{R_{1} R_{2}+R\left(R_{1}+R_{2}\right)}$

Answer 1

Answer: (C)

Induced emf $=B l v . R$ is internal resistance of seat of emf, i.e., of rod
Total resistance of circuit $=R+\frac{R_{1} R_{2}}{R_{1}+R_{2}}$
$\therefore I=\frac{B l v}{R+\frac{R_{1} R_{2}}{\left(R_{1}+R_{2}\right)}}$
$=\frac{B l v\left(R_{1}+R_{2}\right)}{R_{1} R_{2}+R\left(R_{1}+R_{2}\right)}$