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Question

Physics Question on Magnetic Field Due To A Current Element, Biot-Savart Law

Two long parallel wires carry equal current ii flowing in the same direction are at a distance 2d2d apart. The magnetic field BB at a point lying on the perpendicular line joining the wires and at a distance xx from the midpoint is -

A

μ0idπ(d2+x2)\frac{\mu_{0} i d}{\pi\left(d^{2}+x^{2}\right)}

B

μ0ixπ(d2x2)\frac{\mu_{0} i x}{\pi\left(d^{2}-x^{2}\right)}

C

μ0ix(d2+x2)\frac{\mu_{0} i x}{\left(d^{2}+x^{2}\right)}

D

μ0id(d2+x2)\frac{\mu_{0} i d}{\left(d^{2}+x^{2}\right)}

Answer

μ0ixπ(d2x2)\frac{\mu_{0} i x}{\pi\left(d^{2}-x^{2}\right)}

Explanation

Solution

The magnetic field due to two wires at PP
B1=μ0i2π(d+x);B2=μ0i2π(dx)B_{1}=\frac{\mu_{0} i}{2 \pi(d+x)} ; B_{2}=\frac{\mu_{0} i}{2 \pi(d-x)}
Both the magnetic fields act in opposite direction.
B=B2B1\therefore B=B_{2}-B_{1}
=μ0i2π[1dx1ds+x]=\frac{\mu_{0} i}{2 \pi}\left[\frac{1}{d-x}-\frac{1}{d s+x}\right]
=μ0i2π[d+xd+xd2x2]=\frac{\mu_{0} i}{2 \pi}\left[\frac{d+x-d+x}{d^{2}-x^{2}}\right]
=μ0ixπ(d2x2).=\frac{\mu_{0} i x}{\pi\left(d^{2}-x^{2}\right)} .