Question

A straight piece of conducting wire with mass M and length L is placed on a frictionless incline plane tilted at an angle $\theta$ from the horizontal (as shown in fig). There is a uniform, vertical magnetic field at all points (produced by an arrangement of magnets not shown in fig.) To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest :

• A. $\frac{Mgtan\theta }{2LB}$ to the left
• B. $\frac{Mgtan\theta }{LB}$ to the right
• C. $\frac{Mgtan\theta }{LB}$ to the left
• D. $\frac{3Mgtan\theta }{2LB}$ to the left

Answer 1

Answer: (C)

$\frac{Mgtan\theta }{LB}$ to the left

The magnetic force $F=I\overrightarrow{L}\times \overrightarrow{B}$

As wire is in equilibrium, force along and perpendicular to incline should balance each other.

From FBD Diagram of the wire:

$F\cos \theta =Mg\sin \theta$

$ILB\cos \theta =Mg\sin \theta$

$I=\frac{Mg\tan \theta }{LB}$. to the left.