Consider a slowly charging parallel plate capacitor (distance between the plates is d) having circular plates each with an area A, as shown in the figure below. An electric field of magnitude \(E = E_0\sin(\omega t)\) exists between the plates while charging. The associated magnitude of the magnetic field B at the periphery (outer edge) of the capacitor is (Neglect fringe effects)
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A changing electric field acts as a source of magnetic field, described by the displacement current term \(\epsilon_0 \frac{d\Phi_E}{dt}\) in Maxwell's equations. For problems involving a capacitor, this is the key term to use in the Ampere-Maxwell law to find the induced magnetic field.
Step 1: Understanding the Concept:
This problem deals with the magnetic field produced by a changing electric field, a concept central to Maxwell's equations. Specifically, we need to use the Ampere-Maxwell law, which includes the displacement current term responsible for generating a magnetic field from a time-varying electric flux. Step 2: Key Formula or Approach:
The integral form of the Ampere-Maxwell law is:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (I_{enc} + I_d) \]
where \(I_d\) is the displacement current, given by \(I_d = \epsilon_0 \frac{d\Phi_E}{dt}\).
Inside the capacitor, there is no conduction current (\(I_{enc}=0\)). The electric flux is \(\Phi_E = \int \mathbf{E} \cdot d\mathbf{A}\).
We will apply this law to a circular Amperian loop of radius R at the periphery of the capacitor plates. We also use the relation \(c^2 = \frac{1}{\mu_0 \epsilon_0}\). Step 3: Detailed Explanation:
1. Set up the Amperian loop:
Consider a circular Amperian loop of radius R, where R is the radius of the capacitor plates. The periphery of this loop coincides with the outer edge of the capacitor. The area of the plates is \(A = \pi R^2\), so the radius is \(R = \sqrt{A/\pi}\).
2. Apply Ampere-Maxwell Law:
The law states \(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}\).
By symmetry, the magnetic field \(\mathbf{B}\) has a constant magnitude B along the Amperian loop and is tangential to it. The left-hand side (LHS) becomes:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = B \cdot (2\pi R) \]
3. Calculate the electric flux and its time derivative:
The electric field is uniform between the plates and given by \(E = E_0\sin(\omega t)\). The electric flux \(\Phi_E\) through the Amperian loop is:
\[ \Phi_E = E \cdot A = A E_0\sin(\omega t) \]
The time derivative of the electric flux is:
\[ \frac{d\Phi_E}{dt} = \frac{d}{dt} [A E_0\sin(\omega t)] = A E_0\omega\cos(\omega t) \]
4. Equate and solve for B:
Substitute the LHS and the derivative of flux into the Ampere-Maxwell equation:
\[ B \cdot (2\pi R) = \mu_0 \epsilon_0 (A E_0\omega\cos(\omega t)) \]
Using \( \mu_0 \epsilon_0 = \frac{1}{c^2} \):
\[ B \cdot (2\pi R) = \frac{1}{c^2} A E_0\omega\cos(\omega t) \]
Solve for B:
\[ B = \frac{A E_0\omega\cos(\omega t)}{2\pi R c^2} \]
5. Substitute the radius R in terms of area A:
We have \(R = \sqrt{A/\pi}\). Substitute this into the expression for B:
\[ B = \frac{A E_0\omega\cos(\omega t)}{2\pi c^2 \sqrt{A/\pi}} = \frac{\sqrt{A} \sqrt{A} E_0\omega\cos(\omega t)}{2\pi c^2 \sqrt{A}/\sqrt{\pi}} \]
\[ B = \frac{\sqrt{A} \sqrt{\pi} E_0\omega\cos(\omega t)}{2\pi c^2} = \frac{\sqrt{A} E_0\omega\cos(\omega t)}{2\sqrt{\pi} c^2} \]
Rewriting this to match the options:
\[ B = \frac{1}{2c^2} \sqrt{\frac{A}{\pi}} E_0\omega\cos(\omega t) \]
This matches option (A).
Step 4: Final Answer:
The magnitude of the magnetic field B at the periphery of the capacitor is \(\frac{1}{2c^2}\sqrt{\frac{A}{\pi}} E_0\omega\cos(\omega t)\).
