Question

Two parallel rails with negligible resistance are $10.0\text{cm}$ apart. They are connected by a $5.0\Omega$ resistor. The arrangement also contains two metal rods having resistances of $10.0\Omega$ and $15.0\Omega$ as shown in the figure. The rods are pulled away from the resistor at constant speeds of $4.0\text{m/s}$ and $2.0\text{m/s}$ respectively. A uniform magnetic field of magnitude $0.01\text{T}$ is applied perpendicular to the plane of the rails. Determine the current in the $5.0\Omega$ resistor.

• A. $1.45\times {10}^{-4}\text{A}$
• B. $2.90\times {10}^{-4}\text{A}$
• C. $5.25\times {10}^{-4}\text{A}$
• D. $3.25\times {10}^{-4}\text{A}$

Answer 1

The correct option is A $1.45\times {10}^{-4}\text{A}$


Two conductors are moving in the uniform magnetic field, so motional emf will be induced across them.

The rod $ab$ will act as a source of emf,

$e_1=Bvl=\left(0.01\right)\left(4\right)\left(0.1\right)=4\times {10}^{-3}\text{V}$

And its internal resistance is,
$r_1=10.0\Omega$

Similarly, rod $ef$ will also act as source of emf,

$e_2=\left(0.01\right)\times \left(2\right)\times \left(0.1\right)=2\times {10}^{-3}\text{V}$

And its internal resistance is,

$r_2=15.0\Omega$

From right-hand rule, $V_b>V_a$ and $V_e>V_f$.


Also, $R=5.0\Omega$

Now, the equivalent emf of the circuit,

$E_{eq}=\frac{e_1{r}_2-e_2{r}_1}{r_1+r_2}$

$E_{eq}=\frac{60\times {10}^{-3}-20\times {10}^{-3}}{15+10}=1.6\times {10}^{-3}\text{V}$

$R_{eq}$ of the circuit,

$=\frac{15\times 10}{15+10}+5=11\Omega$

$\therefore I=\frac{E_{eq}}{R_{eq}}=\frac{1.6\times {10}^{-3}}{11}$

$\Rightarrow I=1.45\times {10}^{-4}\text{A}$