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 and an equal current flows out of the plane of the page along the shaded portion of the other, as shown. Point A is at the midpoint of the line joining the axes of the conductors.

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Magnetic field at A is $(μ_{0} / 2 π) π d J$ , in the +y direction

Magnetic field at A is $(μ_{0} / 2 π) d^{2} / r,$ in the +y direction

Magnetic field at A is $(μ_{0} / 2 π) 4 d^{2} J / r,$ in the –y direction

Magnetic field inside the cavity is uniform

answer is A, D.

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Detailed Solution

Question Image
$B_{A} = {(B_{A})}_{1} + {(B_{A})}_{2}$

${(B_{A})}_{1} (2 π \frac{d}{2}) = μ_{0} (J π {(\frac{d}{2})}^{2})$

${(B_{A})}_{1} = \frac{μ_{0} J d}{4}$ (points perpendicular to OA, along +y)

${(B_{A})}_{2} (2 π \frac{d}{2}) = μ_{0} (J π {(\frac{d}{2})}^{2})$

${(B_{A})}_{2} = \frac{μ_{0} J d}{4}$ (points perpendicular to O'A, along +y)

Net magnetic field at A

$B_{A} = {(B_{A})}_{1} + {(B_{A})}_{2} = \frac{μ_{0} J d}{4} + \frac{μ_{0} J d}{4} = \frac{μ_{0} J d}{2}$ (along +y) Option (a) is correct. Option (b) is wrong, Option (c) is wrong.

Consider an arbitrary point $P$ which is at a distance $x$ from point $O$ and at a distance $y$ from point $O'$ .

${(B_{P})}_{1} (2 π x) = μ_{0} J (π x^{2})$

${(B_{P})}_{1} = \frac{μ_{0} J x}{2}$

${(B_{P})}_{2} (2 π y) = μ_{0} J (π y^{2})$

${(B_{P})}_{2} = \frac{μ_{0} J y}{2}$

x-components of ${(B_{P})}_{1}$ is ${(B_{P})}_{1} \sin θ_{1}$ $\sin θ_{1} = \frac{l}{x}$
x-components of ${(B_{P})}_{2}$ is ${(B_{P})}_{2} \sin θ_{2}$ $\sin θ_{2} = \frac{l}{y}$
x component of net magnetic field at $P$ is $\frac{μ_{0} J x}{2} \frac{l}{x} - \frac{μ_{0} J y}{2} \frac{l}{y} = 0$
y component of net magnetic field at $P$ is $\frac{μ_{0} J x}{2} \frac{d_{1}}{x} + \frac{μ_{0} J y}{2} \frac{d_{2}}{y} = \frac{μ_{0} J d}{2}$ (along +y)
Magnetic field inside the cavity is uniform. Option (d) is correct.