Question

In figure, wires ${\text{P}}_1{\text{Q}}_1$ & ${\text{P}}_2{\text{Q}}_2$ both are moving towards right with speed $5\text{cm/s}$. Resistance of each wire is $2\Omega$. Then current through $19\Omega$ resistor is : $(B=1\text{T})$

• A. $0$
• B. $0.1\text{mA}$
• C. $0.2\text{mA}$
• D. $0.3\text{mA}$

Answer 1

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

Emf is produced due to the motion of the conductor.


As, emf induced in the conductor,
$E=Blv$

$\Rightarrow E_1=E_2=1\times 4\times {10}^{-2}\times 5\times {10}^{-2}$

$=2\times {10}^{-3}\text{V}=2\text{mV}$


Equivalent resistance,

$R_{eq}=\frac{2\times 2}{2+2}+R=1+19=20\Omega$

Current through $19\Omega$,

$i=\frac{E}{R_{eq}}=\frac{2\times {10}^{-3}}{20}$

$={10}^{-4}\text{A}=0.1\text{mA}$

Hence, $\left(B\right)$ is the correct answer.