Question: 26. A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
(1) zero at all places
(2) constant between the plates and zero outside the plates.
(3) non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates.
(4) zero between the plates and non-zero outside.
Answer: Option (3)
Explanation:
When the capacitor is being charged the electric field E between the plates changes with time because the surface charge density \sigma on the plates changes. The changing electric field produces a displacement current density given by Maxwell’s correction, and this displacement current produces a magnetic field just like a conduction current does.
For a circular plate of radius R, consider a circular Amperian loop of radius
r (coaxial with the plates).
The electric field between the plates (neglecting edge effects) is related to surface charge density
by E=\dfrac{\sigma}{\varepsilon_0}.
Differentiating with respect to time gives
\dfrac{dE}{dt}=\dfrac{1}{\varepsilon_0}\dfrac{d\sigma}{dt}.
The displacement current enclosed by the loop of radius r
is the rate of change of electric flux through the area \pi r^2:
I_{d,\text{enc}}=\varepsilon_0\frac{d}{dt}\left(E\cdot \pi r^2\right)=\varepsilon_0\left(\frac{dE}{dt}\right)\pi r^2.Apply the Ampère–Maxwell law around the circular path
(magnetic field B is tangential and uniform on the path):
\oint \mathbf{B}\cdot d\mathbf{l}=B(2\pi r)=\mu_0 I_{d,\text{enc}}=\mu_0\varepsilon_0\left(\frac{dE}{dt}\right)\pi r^2.Solve for B.
Using \dfrac{dE}{dt}=\dfrac{1}{\varepsilon_0}\dfrac{d\sigma}{dt}
we get for r\le R:
B(r)=\dfrac{\mu_0\varepsilon_0\left(\dfrac{dE}{dt}\right)\pi r^2}{2\pi r}=\dfrac{\mu_0}{2}\left(\dfrac{d\sigma}{dt}\right)r\qquad (r\le R).For r>R the enclosed displacement current
is the total displacement current through the whole plate area \pi R^2, so
B(r)=\dfrac{\mu_0 I_{d,\text{total}}}{2\pi r}=\dfrac{\mu_0}{2\pi r}\left(\dfrac{d\sigma}{dt}\pi R^2\right)=\dfrac{\mu_0}{2}\left(\dfrac{d\sigma}{dt}\right)\dfrac{R^2}{r}\qquad (r>R).Thus B(r) increases linearly with r
inside the plate radius (r\le R)
and falls off as 1/r outside (r>R).
Therefore the magnetic field is non-zero both between and outside
the plates and reaches its maximum value at the edge r=R
(the imaginary cylindrical surface joining the plate peripheries).
Hence option (3) is correct.