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Q.

A long straight wire is fixed horizontally and carries a current of 50 A. A second wire having linear mass density 1.0 × 10-4 kg/m is placed parallel to and directly above this wire at a separation of 5.0 mm. The second wire carries a current [[1]] A in the opposite direction such that the magnetic repulsion can balance its weight.


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Detailed Solution

A long straight wire is fixed horizontally and carries a current of 50 A. A second wire having linear mass density 1.0 × 10-4 kg/m is placed parallel to and directly above this wire at a separation of 5.0 mm. The second wire carries a current of 0.49 A in the opposite direction such that the magnetic repulsion can balance its weight.
Given,
Current through one wire, i1=50 A
Separation between the wires, r=5.0 mm
Converting it to meter, we know that 1 mm= 10-3 m
So, r=5 mm=5 × 10-3 m
Linear mass density of second wire = 1.0 × 10-4 kg/m
Magnetic field is the region in which electric charge experience magnetic influence. A straight wire carrying current will experience a magnetic field due to the current flowing through it.
The magnetic field due to two current-carrying wires is given by B= μ0 i1 i22πr
In order that magnetic repulsion can balance the weight of the wire, B=mg
       μ0 i1 i22πr=mg
       4π × 10-7 ×50 × i22π ×5 × 10-3= 1.0 × 10-4 ×9.8        2 × 10-3 × i2=9.8 × 10-4
       i2= 9.8 × 10-42 × 10-3
       i2=0.49 A
Hence, the current on the second wire is 0.49 A.
 
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