Q.
A straight rod of mass $m\ $and length $L$ is suspended from the identical springs as shown in figure. The spring is stretched a distance $x_{0}$ due to the weight of the wire. The circuit has total resistance$R$. When the magnetic field perpendicular to the plane of paper is switched on, springs are observed to extend further by the same distance. The magnetic field strength is
A) $\frac{2mgR}{LE}$
B) $\frac{mgR}{LE}$
C) $\frac{mgR}{2LE}$
D) $\frac{mgR}{E}$
Explanation
(b) In the absence of magnetic field
$mg = 2kx_{0}$ …..(i)
the current in the rod is $i = \frac{E}{R}\ $
$\therefore\ \ \ \ \ $Magnetic force on the rod is $F_{m} = BiL = \frac{ELB}{R}$
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In downward direction
$\therefore\ \ \ 2kx_{0} = mg + \frac{BLE}{LE}$ …..(ii)
From Eqs. (i) and (ii); we get $4kx_{0} = 2kx_{0} + \frac{BLE}{R}$
$B = \frac{2kx_{0}R}{EL} = \frac{mgR}{LE}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.