JEE Advanced · PhysicsHard

$A parallel plate capacitor with circular plates of radius R is being charged. The displacement current between the plates is equal to the conduction current in the wires. If the rate of change of the electric field between the plates is d E / d t, what is the magnetic field B at a distance r < R from the axis?$

Show correct answer & step-by-step solution

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

Solution

$Apply the Ampere-Maxwell law: \oint B \cdot d l = μ_{0} ϵ_{0} \frac{d}{d t} \int E \cdot d A .$
$For a circular path of radius r < R, the line integral is B (2 π r) and the flux is Φ_{E} = E (π r^{2}) .$
$Substitute these into the law to get B (2 π r) = μ_{0} ϵ_{0} (π r^{2}) \frac{d E}{d t} .$
$Solve for B to obtain B = \frac{μ _{0} ϵ _{0} r}{2} \frac{d E}{d t} .$
$Hence the answer is (A).$