Question

Consider the situation shown in the figure. The wires $P_1{Q}_1$ and $P_2{Q}_2$ are made to slide on the rails with the same speed $5{\text{cm s}}^{-1}$ and have equal resistance of $2\Omega$. Find the electric current in the $19\Omega$ resistor if both the wires move towards right.

• A. $1\text{mA}$
• B. $2\text{mA}$
• C. $0.1\text{mA}$
• D. $0.5\text{mA}$

Answer 1

The correct option is C $0.1\text{mA}$


Conductors are moving in the uniform magnetic field, so motional emf will induce across them.

Both rods will act as a source of emf.

Emf across rod $P_1{Q}_1$ is

${ϵ}_1=Blv=1\times 4\times {10}^{-2}\times 5\times {10}^{-2}=2\times {10}^{-3}\text{V}=2\text{mV}$

Rod $P_2{Q}_2$ is exactly the same as rod $P_1{Q}_1$ and it is also moving with the same speed in the same direction.

Therefore, emf across rod $P_2{Q}_2$ is also $2\text{mV}$.

From right-hand rule, we can say that,

$V_{P_1}>V_{Q_1}$

$V_{P_2}>V_{Q_2}$

Therefore, the equivalent circuit can be drawn as -


The equivalent emf of the circuit is,

$E_{eq}=\frac{{ϵ}_1{r}_2+{ϵ}_2{r}_1}{r_1+r_2}$

$\Rightarrow E_{eq}=\frac{2\times 2+2\times 2}{2+2}=2\text{mV}$

$r_{e}q$ of the circuit,

$r_{eq}=\frac{2\times 2}{2+2}=1\Omega$

$\therefore$ Current flowing through $19\Omega$ resistance is,

$I=\frac{E_{eq}}{r_{eq}+R}$

$\Rightarrow I=\frac{2\times {10}^{-3}}{1+19}=0.1\text{mA}$