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Physics-

General

Easy

Question

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 $capital omega$ and 5 $capital omega$ are to slide without friction along the rail. The distance between the conducting rails is 0.1 m. Then

Induced current $= \frac{1}{150} A$ directed clockwise if 10 W resistor is pulled to the right with speed 0.5 ms^–1 and $5 Ω$ resistor is held fixed
Induced current $= \frac{1}{300} A$ directed anti-clockwise if 10 W resistor is pulled to the right with speed 0.5 ms^–1 and $5 Ω$ resistor is held fixed
Induced current $= \frac{1}{300} A$ directed clockwise if 5 W resistor is pulled to the left at 0.5 ms^–1 and 10 W resistor is held at rest
Induced current $= \frac{1}{150} A$ directed anti-clockwise if 5 W resistor is pulled to the left at 0.5 ms^–1 and 10 W resistor is held at rest

The correct answer is: Induced current $equals fraction numerator 1 over denominator 150 end fraction A$ directed anti-clockwise if 5 W resistor is pulled to the left at 0.5 ms^–1 and 10 W resistor is held at rest

When $5 capital omega$ resistor is pulled left at 0.5 m/sec induced emf., in the said resistor = $e equals v B l equals 0.5 cross times 2 cross times 0.1 equals 0.1 V$
Resistor $10 capital omega$ is at rest so induced emf in it $left parenthesis e equals v B l right parenthesis$ be zero.

Now net emf.,
in the circuit $equals 0.1 V$
and equivalent
resistance of the circuit
R = 15 W
Hence current $i equals fraction numerator 0.1 over denominator 15 end fraction a m p equals fraction numerator 1 over denominator 150 end fraction a m p$
And its direction will be anti-clockwise (according to Lenz’s law)

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