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Question

Two long straight parallel conductors carry steady current I1 and I2 separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other.
Obtain the expression for this force. Hence define one ampere.

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Solution

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 and when they carry currents in opposite directions, they experience a repulsive force.
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 DeltaL of wire RS. The magnetic field produce by current carrying conductor PQ at the location of other wire RS
B1=μ0I12πr.....(i)
According to maxwell's right hand rule or right hand palm rule number 1, the direction of B1 will be perpendicular to the plane of paper and direct 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
ΔF=B1I2 ΔLsin90o=μ0I12πrI2ΔL
The total force on conductor of length L will be
F=μ0I1I22πrΔL=μ0I1I22πrL
Force acting per unit length of conductor
f=FL=μ0I1I22πrN/m....(ii)
According to Fleming's left hand rule, the direction of magnetic force will be towards PQ i.r, the force will be attractive.
On the other hand if the currents I1 and I2 wires are in opposite direction, the force will be repulsive. The magnitude of force in each case remains the same.
Definition of SI unit of Current (ampere): In SI system of fundamental unit of current 'ampere ' is defined assuming the force between the two current carrying wires as standard.
The force between two parallel current carrying conductors r is
f=FL=μ0I1I22πrN/m
If I1=i2=1 A,r=1 m, then
f=μ02π=2×102N/m
Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2×107 on 1 m length of either wire.


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