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A straight rod of mass m and length L is suspended from the identical spring as shown in the figure. The spring stretched by a distance of $x_{0}$ due to the weight of the wire. The circuit has total resistance $R Ω$ . When the magnetic field perpendicular to the plane of the paper is switched on, springs are observed to extend further by the same distance. The magnetic field strength is

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\frac{mgR}{ε x_{0}}

; directed into the plane of the paper

\frac{mgR}{ε L}

; directed into the plane of the paper

\frac{mgR}{2 ε x_{0}}

; directed outward from the plane of the paper

\frac{mgR}{ε L}

; directed outward from the plane of the paper

answer is A.

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Detailed Solution

In the absence of magnetic field

$mg = 2 k x_{0} \to (1)$

The current in the rod is $I = \frac{ε}{R}$

Magnetic force on the rod is

$F_{M} = B i L = \frac{ε LB}{R}$ Downward direction

$4 k x_{0} = mg + \frac{ε LB}{R} \to (2)$

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From (1) and (2)

$4 k x_{0} = 2 k x_{0} + \frac{ε LB}{R}$

$2 k x_{0} = \frac{ε LB}{R}$

$\Rightarrow B = \frac{2 k x_{0} R}{ε L} = \frac{mgk}{ε L}$

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