Magnetic fields

Discharging a Capacitor – Magnetic Field Aspects

When a charged capacitor is connected across a resistor, the stored electric energy is released as a current. Although the current is transient, the changing electric field in the capacitor gives rise to a magnetic field. This phenomenon is a direct illustration of the Maxwell–Ampère law and is essential for understanding electromagnetic induction at A‑Level.

1. The RC Discharge Circuit

The basic circuit consists of a capacitor $C$ initially charged to a voltage $V_0$, a resistor $R$, and a switch that closes at $t=0$ to start the discharge.

2. Governing Equations

• Kirchhoff’s loop rule gives the differential equation for the charge $Q(t)$ on the capacitor: $$\frac{dQ}{dt} + \frac{Q}{RC}=0$$
• Solution for charge and current: $$Q(t)=Q_0 e^{-t/RC},\qquad I(t)=\frac{dQ}{dt}=-\frac{Q_0}{RC}e^{-t/RC}$$ where $Q_0 = C V_0$.
• Electric field between the plates: $$E(t)=\frac{V(t)}{d}=\frac{Q(t)}{C d}$$ with $d$ the plate separation.
• Displacement current density (Maxwell’s addition): $$J_D=\varepsilon_0\frac{\partial E}{\partial t}$$

3. Magnetic Field Produced During Discharge

According to the Maxwell–Ampère law, the line integral of the magnetic field $\mathbf{B}$ around a closed loop equals the sum of the conduction current $I$ and the displacement current $I_D$ passing through the surface bounded by the loop:

$$\oint \mathbf{B}\cdot d\mathbf{l}= \mu_0\left(I + \varepsilon_0\frac{d\Phi_E}{dt}\right)$$

For the RC discharge, the conduction current in the external circuit and the displacement current between the plates are equal at any instant:

$$I_D = \varepsilon_0\frac{d\Phi_E}{dt}=I(t)$$

Thus the magnetic field outside the capacitor is the same as that produced by a real current $I(t)$ flowing through the connecting wires.

4. Magnetic Field Around a Straight Wire

Using Ampère’s law for a circular path of radius $r$ around the wire:

$$\oint \mathbf{B}\cdot d\mathbf{l}=B(2\pi r)=\mu_0 I(t)$$

Hence the magnitude of the magnetic field at distance $r$ from the wire is:

$$B(r,t)=\frac{\mu_0 I(t)}{2\pi r}= \frac{\mu_0 Q_0}{2\pi r RC}e^{-t/RC}$$

5. Magnetic Field Inside the Capacitor

Consider a circular Amperian loop of radius $r$ ( $r $$I_D(r)=I(t)\frac{r^2}{a^2}$$

Applying the Maxwell–Ampère law gives:

$$B(r,t)=\frac{\mu_0 I(t)}{2\pi a^2}r \qquad (r\le a)$$

This linear dependence on $r$ is a key result that can be tested experimentally.

6. Summary Table of Key Relations

Quantity	Expression	Notes
Charge on capacitor	$Q(t)=Q_0 e^{-t/RC}$	Exponential decay
Current in circuit	$I(t)=-\dfrac{Q_0}{RC}e^{-t/RC}$	Direction opposite to charge flow
Displacement current	$I_D(t)=\varepsilon_0\frac{d\Phi_E}{dt}=I(t)$	Equal to conduction current
Magnetic field outside wire	$B(r,t)=\dfrac{\mu_0 I(t)}{2\pi r}$	Same as a steady current at that instant
Magnetic field inside capacitor (r ≤ a)	$B(r,t)=\dfrac{\mu_0 I(t)}{2\pi a^2}r$	Linear with radius

7. Example Calculation

1. Given: $C=10\;\mu\text{F}$, $R=5\;\text{k}\Omega$, $V_0=12\;\text{V}$, plate radius $a=2\;\text{cm}$.
2. Find the magnetic field $B$ at $r=1\;\text{cm}$ inside the capacitor after $t=2\tau$, where $\tau=RC$.
3. Solution:
   • Time constant $\tau = RC = (5\times10^3)(10\times10^{-6}) = 0.05\;\text{s}$.
   • At $t=2\tau = 0.10\;\text{s}$, $$I(t)=\frac{V_0}{R}e^{-t/\tau}= \frac{12}{5\times10^3}e^{-2}=2.4\times10^{-3}e^{-2}\;\text{A}\approx3.25\times10^{-4}\;\text{A}$$
   • Magnetic field inside: $$B(r,t)=\frac{\mu_0 I(t)}{2\pi a^2}r =\frac{4\pi\times10^{-7}\times3.25\times10^{-4}}{2\pi (0.02)^2}(0.01) \approx 1.6\times10^{-9}\;\text{T}$$

8. Common Misconceptions

• “No magnetic field because the current is decreasing.” – The magnetic field depends on the instantaneous current, not on whether it is increasing or decreasing.
• “Displacement current is a fictitious quantity.” – It is a real term required for continuity of the magnetic field in regions where there is no conduction current.
• “Magnetic field inside the capacitor is zero because there are no wires.” – The changing electric field produces a displacement current, which generates a magnetic field just as a wire does.

9. Practice Questions

1. Derive the expression for the magnetic field at a distance $r$ from the centre of a circular capacitor plate during discharge, assuming uniform electric field between the plates.
2. A capacitor of $22\;\mu\text{F}$ is discharged through a $10\;\text{k}\Omega$ resistor. Calculate the magnetic field at $r=5\;\text{mm}$ from the wire after one time constant.
3. Explain qualitatively how the direction of the magnetic field changes if the polarity of the initially charged capacitor is reversed before discharge.

10. Further Reading (Suggested Topics)

• Maxwell’s equations and the role of displacement current.
• Induced emf in a loop surrounding a discharging capacitor.
• Energy stored in magnetic fields: $U_B = \frac{1}{2\mu_0}\int B^2\,dV$.