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A pair of parallel conducting rails lie at right angle to a uniform magnetic field of 2.0 T as shown in the fig. Two resistors 10 Ω and 5 Ω are to slide without friction along the rail. The distance between the conducting rails is 0.1 m. Then

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Induced current $= \frac{1}{{150}}A$ directed clockwise if 10 Ω resistor is pulled to the right with speed 0.5 ms^–1 and 5Ω resistor is held fixed

Induced current $= \frac{1}{{300}}A$ directed anti-clockwise if 10 Ω resistor is pulled to the right with speed 0.5 ms^–1 and 5Ω resistor is held fixed

Induced current $= \frac{1}{{300}}A$ directed clockwise if 5 Ω resistor is pulled to the left at 0.5 ms^–1 and 10 Ω resistor is held at rest

Induced current $= \frac{1}{{150}}A$ directed anti-clockwise if 5 Ω resistor is pulled to the left at 0.5 ms^–1 and 10 Ω resistor is held at rest

answer is D.

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Detailed Solution

When 5Ω 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 10Ω is at rest so induced emf in it

be zero.
Now net emf.,
in the circuit = 0.1 V
and equivalent
resistance of the circuit
R = 15 Ω
Hence current

i = \frac{{0.1}}{{15}}amp = \frac{1}{{150}}amp

And its direction will be anti-clockwise (according to Lenz’s law)

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