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Question

A conductor PQ of length b carries a current i2 is placed perpendicular to a long conductor XY carrying a current i1 as shown in figure. The magnitude and direction of the magnetic force acting on PQ is

Both the conductors are in xyplane.


A
μ0i1i24πln(a+ba)^j
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B
μ0i1i22πln(a+ba)^j
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C
μ0i1i24πln(a+ba)^j
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D
μ0i1i22πln(a+ba)^j
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Solution

The correct option is D μ0i1i22πln(a+ba)^j

Let us consider, a current element of size dx on PQ at a distance x from the current carrying conductor PQ as shown in figure.


Force on the current element,

dF=i2(dx×B)

=i2[dx^i×μ0i12πx(^k)] [B:is the field due to XY]

=μ0i1i2dx2πx^j

Integrating, both sides with proper limit,

F0dF=μ0i1i22πa+ba1xdx

F=μ0i1i22πln(a+ba)

Since, the direction of the force is along +^j direction,

F=μ0i1i22πln(a+ba)^j

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (D) is the correct answer.
Why this question?

To understand the concept of force between two straight current carrying conductors.



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