Question

Obtain an expression for the force between two current carrying loops and show that it obeys Newton’s Third Law.

Answer 1

Answer:

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Explanation:

Force per unit length between two long straight parallel conductors: Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I1  and I2 respectively. It has been observed experimentally that when the currents in the wire are in the same direction, they experience an attractive force (fig. a) and when they carry currents in opposite directions, they experience a repulsive force (fig. b). Let the conductors PQ and RS carry currents I1  and I2 in same direction and placed at separation r.

Consider a current–element ‘ab’ of length ΔL of wire RS. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS

According to Maxwell’s right hand rule or right hand palm rule no. 1, the direction of B1 will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length ΔL