Question

A capacitor is made of two circular metal plates of radius 1 m, separated by a distance \(d = 1\text{ mm}\). The space between the plates is filled with a dielectric of permittivity \(\epsilon_r = 5\). The capacitor is connected to a voltage \(V = 10\sin(2\pi \times 10^6 \times t)\) volts, where \(t\) is in seconds. The maximum value of the magnetic field between the plates, at a radial distance \(r = 0.5\text{ m}\) from the centre, is \(B \times 10^{-6}\text{ T}\). The value of \(B\) (rounded off to two decimal places) is ______.
(Speed of light in vacuum \(c = 3 \times 10^8\text{ m.s}^{-1}\))

Hint:

Hint:
Use the Ampere-Maxwell law with the displacement current between the plates: \(B = \dfrac{\epsilon_r r}{2dc^2}\dfrac{dV}{dt}\), and use the maximum rate of change of the given sinusoidal voltage.

Answer 1

Correct Answer: 0.87

Step 1: Set up the electric field between the plates.
Treat the capacitor as an ideal parallel plate capacitor with a uniform field $E(t) = V(t)/d$ between the plates. This is a fair approximation near the axis, well inside the plate edges.

Step 2: Find the enclosed displacement current.
The displacement current density is $J_d = \epsilon \, dE/dt$, where $\epsilon = \epsilon_r \epsilon_0$. For an Amperian loop of radius $r$ centred on the axis, the enclosed displacement current is:
\[ I_{d,enc} = \epsilon \frac{dE}{dt}(\pi r^2) = \epsilon_r \epsilon_0 \frac{r^2}{d}\frac{dV}{dt}\pi \]

Step 3: Apply Ampere's law to get B.
Ampere's law gives $B(2\pi r) = \mu_0 I_{d,enc}$, so:
\[ B = \frac{\mu_0 I_{d,enc}}{2\pi r} = \frac{\mu_0 \epsilon_r \epsilon_0 \, r}{2d}\frac{dV}{dt} \]
Using $\mu_0\epsilon_0 = 1/c^2$ (with the value of $c$ given in the question) avoids needing $\epsilon_0$ separately:
\[ B = \frac{\epsilon_r r}{2dc^2}\frac{dV}{dt} \]

Step 4: Substitute the peak rate of change of voltage.
$V(t) = 10\sin(\omega t)$ with $\omega = 2\pi \times 10^6$ rad/s, so the largest value $dV/dt$ can take is $10\omega = 2\pi \times 10^7$ V/s. Plugging in $\epsilon_r = 5$, $r = 0.5$ m, $d = 10^{-3}$ m and $c = 3\times 10^8$ m/s:
\[ B_{max} = \frac{5 \times 0.5}{2\times 10^{-3}\times (3\times 10^8)^2}\times 2\pi\times 10^7 = 8.73\times 10^{-7}\text{ T} \]

Final Answer:
Written in the form asked in the question, $B_{max} = 0.87 \times 10^{-6}$ T. \[ \boxed{B = 0.87} \]