Question bank/JEE Main/Physics

Two long conductors are arranged as shown above to form
overlapping cylinders, each of radius r,whose centers are separated by a
distance d. Current of density J flows into the plane of the page along
the shaded part of one conductor andan equal current flows out of the
plane of the page along the shaded portion of the other, as shown. What
are the magnitude and direction of themagnetic field at point A?

A$\left( \frac{\mu_{0}}{2\pi} \right)\pi dJ,\$in the +y- directionB$\left( \frac{\mu_{0}}{2\pi} \right)\frac{d^{2}}{r},$ in
the +y directionC$\left( \frac{\mu_{0}}{2\pi} \right)\frac{4d^{2}J}{r},$ in
the -y directionD$\left( \frac{\mu_{0}}{2\pi} \right)\frac{{Jr}^{2}}{d},$ in
the -y direction

Show answer and solution▾

[IMAGE] Fill the over-lapping part with +J and -J current density
Using principle of super position
Magnetic field due to conductor
$B_{M}(2\pi OA) = \mu_{0}J\pi(OA)^{2}$
$B_{M} = \frac{\mu_{0}J}{2\pi}\pi(OA)$ (in y-direction)
And magnetic field due to conductor
$B_{N}\left( 2\pi O^{'}A \right) = \mu_{0}J\pi\left( O^{'}A \right)^{2}$
$B_{N} = \frac{\mu_{0}J}{2\pi}\pi\left( O^{'}A \right)$ in y-direction 
So Net is given by
$B_{Net} = B_{m} + B_{N} = \frac{\mu_{0}J}{2\pi}\pi\left\lbrack OA + O^{'}A \right\rbrack$
$= \frac{\mu_{0}J\pi d}{2\pi}$ in y -direction

Source: Grass-G / P-13 Physics paper 4 Nov.docx