QJEEAdv18Ph8ba8: JEE Advanced 2018 Physics - Class 12 JEEAdv18Ph8ba8 Solution: Two infinitely long straight wires lie in the -plane along the…

The magnetic field at the origin O (0, 0, 0) due to the infinitely long straight wire located at x = +R is

${\overrightarrow B _1}(O) = {{{\mu _0}{i_1}} \over {2\pi R}}\widehat k$

and due to the infinitely long straight wire located at x = $-$ R is

${\overrightarrow B _2}(O) = {{{\mu _0}{i_2}} \over {2\pi R}}( - \widehat k)$ .

The magnetic field at O due to the circular loop is

${\overrightarrow B _3}(O) = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {{(\sqrt 3 R)}^2})}^{3/2}}}}( - \widehat k) = {{{\mu _0}i} \over {16R}}( - \widehat k)$ .

The resultant magnetic field at the origin (0, 0, 0) is

${\overrightarrow B _3}(O) = {\overrightarrow B _1}(O) + {\overrightarrow B _2}(O) + {\overrightarrow B _3}(O)$

$= {{{\mu _0}} \over {16\pi R}}(8({i_1} - {i_2}) - \pi i)\widehat k$ .

Thus, if i₁ = i₂ then ${\overrightarrow B _1}(O) = - {\overrightarrow B _2}(O)$ and ${\overrightarrow B _1}(O) = {\overrightarrow B _3}(O) \ne \overrightarrow 0$ .

If i₁ > 0 and i₂ < 0 then both ${\overrightarrow B _1}(O)$ and ${\overrightarrow B _2}(O)$ are along $\widehat k$ but ${\overrightarrow B _3}(O)$ is along $-$ $\widehat k$ . Thus, resultant field $\overrightarrow B (O)$ can be zero.

If i₁ < 0 and i₂ > 0 then ${\overrightarrow B _1}(O)$ , ${\overrightarrow B _2}(O)$ and ${\overrightarrow B _3}(O)$ are all along $-$ $\widehat k$ . Thus, resultant field $\overrightarrow B (O)$ cannot be zero. We encourage you to relate these results with the feasible solutions of 8(i₁ $-$ i₂) $-$ $\pi$ i = 0 for i > 0.

Now, let us find the magnetic field at the centre of the loop i.e., point C in the figure.

The magnetic field $\overrightarrow B_1$ at the point C due to the infinitely long straight wire located x = +R is perpendicular to the line AC (see figure). It lies in the x-z plane and makes an angle 30 $^\circ$ with the x axis. Resolve $\overrightarrow B_1$ along the x and z directions to get

${{\vec B}_1}(C) = {{{\mu _0}{i_1}} \over {2\pi (2R)}}(\cos 30^\circ \widehat i + \sin 30^\circ \widehat k)$ .

Similarly, the magnetic field at C due to the infinitely long straight wire located at x = $-$ R is

${{\vec B}_2}(C) = {{{\mu _0}{i_2}} \over {2\pi (2R)}}(\cos 30^\circ \widehat i - \sin 30^\circ \widehat k)$ .

The magnetic field at C due to the circular loop is

${{\vec B}_3}(C) = {{{\mu _0}i} \over {(2R)}}( - \widehat k)$ .

The resultant magnetic field at C is

${{\vec B}_1}(C) = {{\vec B}_1}(C) + {{\vec B}_2}(C) + {{\vec B}_3}(C) = {{{\mu _0}} \over {8\pi R}}\left( {\sqrt 3 ({i_1} + {i_2})\widehat i + ({i_1} - {i_2} - 4\pi i)\widehat k} \right)$ .

If i₁ = i₂ then magnitudes of ${{\vec B}_1}(C)$ and ${{\vec B}_2}(C)$ are equal and their z-components are equal in magnitude but opposite in direction. Thus, z-component of the magnetic field at C is due to the circular loop only and its value is $\left( { - {{{\mu _0}i} \over {2R}}} \right)$ .