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: Understanding the Concept:
A charging or discharging capacitor has no real current flowing between its plates, but the changing electric field there acts as a displacement current. By the Ampere-Maxwell law, this displacement current produces a magnetic field between the plates, just as a real current would in a wire.

Step 2: Key Formula or Approach:
For a parallel plate capacitor of plate separation \(d\), the field between the plates is \(E = V/d\). Applying the Ampere-Maxwell law on a circular loop of radius \(r\) (less than the plate radius) centred on the axis:
\[ B (2\pi r) = \mu_0 \epsilon \frac{dE}{dt}(\pi r^2) \]
where \(\epsilon = \epsilon_r \epsilon_0\) is the permittivity of the dielectric. This gives:
\[ B = \frac{\mu_0 \epsilon_r \epsilon_0 \, r}{2d}\frac{dV}{dt} \]
Since \(\mu_0 \epsilon_0 = 1/c^2\), this can be written using only the given constants as:
\[ B = \frac{\epsilon_r \, r}{2 d c^2}\frac{dV}{dt} \]

Step 3: Detailed Explanation:
Differentiate the applied voltage: \(V = 10\sin(2\pi \times 10^6 t)\), so \(dV/dt = 10 (2\pi \times 10^6)\cos(2\pi \times 10^6 t)\). The maximum value of this rate is:
\[ \left(\frac{dV}{dt}\right)_{max} = 10 \times 2\pi \times 10^6 = 2\pi \times 10^7 \text{ V/s} \]
Now put in the numbers, with \(\epsilon_r = 5\), \(r = 0.5\text{ m}\), \(d = 10^{-3}\text{ m}\), \(c^2 = 9 \times 10^{16}\text{ m}^2\text{s}^{-2}\):
\[ B_{max} = \frac{5 \times 0.5}{2 \times 10^{-3} \times 9 \times 10^{16}} \times 2\pi \times 10^7 = \frac{2.5}{1.8 \times 10^{14}} \times 2\pi \times 10^7 \]
\[ B_{max} = 8.73 \times 10^{-7}\text{ T} = 0.87 \times 10^{-6}\text{ T} \]

Final Answer:
So \(B = 0.87\), since \(B_{max} = 0.87 \times 10^{-6}\text{ T}\). \[ \boxed{B = 0.87} \]