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Question

An arrangement of three infinitely long straight wires placed perpendicular to plane of paper carrying same current I along the same direction is shown in the figure. The magnitude of force per unit length on wire B is


A
μ0I22πd
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B
2μ0I2πd
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C
μ0I22πd
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D
2μ0I2πd
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Solution

The correct option is C μ0I22πd
Clearly, the magnetic field due to wire C ie BC is pointing downwards as shown in the figure.

Hence, the force on wire B due to wire C ie FC is pointing rightwards as shown below.

Similarly, the magnetic field due to wire A ie BA is pointing leftwards as shown in the figure.

Hence, the force on wire B due to wire A ie FA is pointing downwards as shown in the figure.


|FA|l=|FC|l=μ0I22πd=f(say)

Fnetl=f2+f2+2f.f.cos90=f2

Fnetl=μ0I222πd=μ0I22.πd
Why this Question?
Note: The magnitude of force per unit length between two parallel current carrying infinite wires having currents I1 and I2 is given by,

F=μ0I1I22πd

Where, d = seperation between the two parallel wires.

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