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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 xyâˆ’plane.

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=μ0i1i22π∫a+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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