JEE Main · PhysicsHard

$A parallel plate capacitor with circular plates of radius R is being charged. The displacement current density between the plates is uniform and equal to J_{d} . The magnetic field B at a distance r from the axis of the plates (where r < R) is:$

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Correct answer: A — $\frac{μ _{0} J _{d} r}{2}$

Solution

$Apply the Ampere-Maxwell law: \oint B \cdot d l = μ_{0} I_{d} .$
$The displacement current enclosed by a loop of radius r is I_{d} = J_{d} (π r^{2}) .$
$The line integral of the magnetic field is B (2 π r) .$
$Equating the two: B (2 π r) = μ_{0} J_{d} (π r^{2}) .$
$Solving for B gives B = \frac{μ _{0} J _{d} r}{2} .$
$Hence the answer is (A).$