First slide

Force on a current carrying wire placed in a magnetic field

Question

An infinitely long, straight conductor AB is fixed and a current is passed through it. Another movable straight wire CD of finite length and carrying current is held perpendicular to it and released. Neglect weight of the wire

Moderate

A

The rod CD will move upwards parallel to itself
B

The rod CD will move downward parallel to itself
C

The rod CD will move upward and turn clockwise at the same time
D

The rod CD will move upward and turn anticlockwise at the same time

Solution

The magnetic field (produced by AB) at different points of CD are different and gradually decreases as we are moving from end C to end D ,as a result force per unit length on CD gradually decreases as we are moving from end C to end D .Therefore the the resultant force on CD acts at a point which lies to the left of the centre of mass of CD .This resultant force produces a clockwise moment about the centre of mass of CD'. Hence CD will experience force and torque. From Flemming"s left hand rule it can be seen that the force will be upward .

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