Wires 1 and 2 carrying currents i_1 and i_2 respectively are inclined at an angle \theta to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1 ?

\mu _0 \over 2\pi ri_1i_2\,dl\,\cos \,\theta

\mu _0 \over 2\pi ri_1i_2\,dl\,\tan \,\theta

\mu _0 \over 2\pi ri_1i_2\,dl\,\sin \,\theta

\mu _0 \over 4\pi ri_1i_2\,dl\,\sin \,\theta

Magnetics PYQ — JEE Main 2002 · Shift 0 · Q126 with Solution

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Correct answer: C

Magnetic field due to wire 1
For a long straight wire carrying current $i_1$ , the magnetic field at a perpendicular distance $r$ is

B_1=\frac{\mu_0 i_1}{2\pi r}.

Force on a current element of wire 2
A small element $d\ell$ of wire 2 carrying current $i_2$ experiences magnetic force

d\vec F=i_2\, d\vec \ell \times \vec B.

Hence magnitude is

dF=i_2\, d\ell\, B_1\sin\phi,

where $\phi$ is the angle between $d\vec\ell$ and $\vec B$ .

Determine the angle between wire 2 and magnetic field
The magnetic field due to wire 1 is circular around wire 1 and lies perpendicular to wire 1.
If wire 2 makes an angle $\theta$ with wire 1, then the angle between wire 2 and the magnetic field is

\phi=90^\circ-\theta.

Therefore,

\sin\phi=\sin(90^\circ-\theta)=\cos\theta.

dF=i_2 d\ell \left(\frac{\mu_0 i_1}{2\pi r}\right)\cos\theta.

Thus,

\boxed{dF=\frac{\mu_0}{2\pi r} i_1 i_2 \, d\ell\, \cos\theta}

Match with options
This corresponds to Option C.
Comparison with stored answer
Stored correct answer: C
Derived answer: C
So they agree.