Q. A pair of parallel conducting rails lie at right angles to a uniform magnetic field of $2.0\ T$ as shown in the fig. Two resistors $10\ \Omega$ and $5\ \Omega$ are to slide without friction along the rail. The distance between the conducting rails is $0.1\ m$. Then
A) Induced current $= \frac{1}{150}A$ directed clockwise if $10\ \Omega$ resistor is pulled to the right with speed $0.5\ ms^{- 1}$ and $5\Omega$ resistor is held fixed
B) Induced current $= \frac{1}{300}A$ directed anti-clockwise if $10\ \Omega$ resistor is pulled to the right with speed $0.5\ ms^{- 1}$ and $5\Omega$ resistor is held fixed
C) Induced current $= \frac{1}{300}A$ directed clockwise if $5\ \Omega$ resistor is pulled to the left at $0.5\ ms^{- 1}$ and $10\Omega$ resistor is held at rest
D) Induced current $= \frac{1}{150}A$ directed anti-clockwise if $5\ \Omega$ resistor is pulled to the left at $0.5\ ms^{- 1}$ and $10\Omega$ resistor is held at rest
Explanation
(d) When $5\Omega$ resistor is pulled left at $0.5\ m/sec$ induced $emf$ in the said resistor $= e = vBl = 0.5 \times 2 \times 0.1 = 0.1\ V$
Resistor $10\Omega$ is at rest so induced $emf$ in it $(e = vBl)$ be zero
Now net $emf$ in the circuit $= 0.1V$ and equivalent resistance of the circuit $R = 15\Omega$
Hence current $i = \frac{0.1}{15}amp = \frac{1}{150}amp$
And its direction will be anti-clockwise (according to Lenz’s law)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.