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The force per unit length between two pa...

The force per unit length between two parallel current carrying wires `= (mu_(0)i_(1)i_(2))/(2pir)` . The force is attractive when the current is in same direction and repulsive, when the they are in opposite directions. The force between the wires of two parallel wires is shown. We can determine the equilibrium position. Then we displace upper wire by a small distance, keeping lower wire fixed. If the wire returns to or tries to return to its equilibrium position, its equilibrium is stable. We can thus show that upper wire can execute linear simple harmonic motion or not. The length of wire AB is large as compared to separation between the wires.

If `lamda` is mass per unit length of wire CD, then the equilibrium separation h is given by

A

`h = (mu_(0)i_(1) i_(2))/(2pi lamdag)`

B

`h = (2mu_(0)i_(1) i_(2))/(2pi lamdag)`

C

`h = (2pi lamdag)/(mu_(0)i_(1)i_(2))`

D

`h = (4pi lamda g)/(mu_(0)i_(1) i_(2))`

Text Solution

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The correct Answer is:
A
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The force per unit length between two parallel current carrying wires = (mu_(0)i_(1)i_(2))/(2pir) . The force is attractive when the current is in same direction and repulsive, when the they are in opposite directions. The force between the wires of two parallel wires is shown. We can determine the equilibrium position. Then we displace upper wire by a small distance, keeping lower wire fixed. If the wire returns to or tries to return to its equilibrium position, its equilibrium is stable. We can thus show that upper wire can execute linear simple harmonic motion or not. The length of wire AB is large as compared to separation between the wires. If wire CD is displaced upward to increase the separation by dh, the magnitude of net force per unit length acting on the wire CD becomes

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