Question

Figure shows a wire $ab$ of length $l$ and resistance $R$ which can slide on a smooth pair of rails in a uniform magnetic field $B$. $I_g$ is a current generator which supplies a constant current $i$ in the circuit. If the wire $ab$ slides at a speed $v$ towards the right, find the potential difference $V_a-V_b$.

• A. $vBl+iR$
• B. $vBl-iR$
• C. $iR$
• D. $0$

Answer 1

The correct option is B $vBl-iR$

The moving wire $ab$ is equivalent to a battery of emf $vBl$ having a resistance $R$.

From the right-hand rule, the positive terminal of the equivalent battery is towards $a$.

An equivalent circuit is shown in the figure.

The potential difference between $a$ and $b$ is,
$V_a-V_b=vBl-iR$.