Question

A pair of parallel conducting rails lies at right angle to a uniform magnetic field of $2.0\, T$ as shown in the figure. 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 $10 \Omega$ 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 $10 \Omega$ 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 5A resistor is pulled to the left at $5\, ms ^{-1}$ and $10 \Omega$ resistor is held at rest

Answer 1

Answer: (D)

When $5\, \Omega$ resistor is pulled left at $0.5\, m / s$, induced emf in the said resistor $=e=v B l=0.5 \times 2 \times 0.1=0.1\, V$ Resistor $10 \Omega$ is at rest so induced emf in it $(e=v B l$ ) is zero.

Now net emf in the circuit $=0.1\, V$
and equivalent resistance of the circutit $R=15\, \Omega$
Hence, current, $i=\frac{0.1}{15} A =\frac{1}{150} A$
And its direction will be anti-clockwise (according to Lenz's law)