Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.1.3

Rotating charged disk

Let the disk $S = {(x, y, z) : x^{2} + y^{2} \leq b^{2}, z = 0}$ carry a static surface charge distributed uniformly with a density of $ρ_{S}$ and be rotated about the $z$ -axis with constant angular velocity $ω$ . (a) Write the density of the corresponding current. (b) Find the magnetic field created at the point $P (0, 0, h)$ .

Source: B. M. Notaroš, Electromagnetics, Problem 4.5.

Try it yourself first.

Strategy

A charge that moves is a current. A point of the disk at radius $ρ^{'}$ moves on a circle with speed $ω ρ^{'}$ in the $a_{ϕ}$ direction, so the disk is a nest of concentric current loops, all coaxial with the $z$ -axis. P lies on that axis, and the on-axis field of one loop is known from Problem 4.1.2: integrate it over the rings.

1. Symmetry

The disk is invariant under any rotation about $z$ , and P is on the axis: only $B_{z}$ can survive at P (Problem 4.1.2).

2. Current density (part a)

The velocity of the charge at $(ρ^{'}, ϕ^{'}, 0)$ is $v = ω ρ^{'} a_{ϕ}$ , hence the surface current density is

Answer

J_{S} = ρ_{S} v = ρ_{S} ω ρ^{'} a_{ϕ} [A/m], 0 \leq ρ^{'} \leq b

As a volume density, $J = ρ_{S} ω ρ^{'} δ (z^{'}) a_{ϕ}$ . The current crossing a radial segment of length $d ρ^{'}$ is $d I = ∣ J_{S} ∣ d ρ^{'} = ρ_{S} ω ρ^{'} d ρ^{'}$ : this is the current of the ring between $ρ^{'}$ and $ρ^{'} + d ρ^{'}$ (Figure 4.1.20a). Moving charge is current, and the disk is a nest of concentric loops.

3. The field of one ring

Each ring is a loop of radius $ρ^{'}$ carrying $d I$ , and by Problem 4.1.2 its field at P is along $a_{z}$ :

d B_{z} = \frac{μ _{0} d I ρ ^{' 2}}{2 ( ρ ^{' 2} + h ^{2} ) ^{3/2}} = \frac{μ _{0} ρ _{S} ω}{2} \frac{ρ ^{' 3} d ρ ^{'}}{( ρ ^{' 2} + h ^{2} ) ^{3/2}} .

4. Integrate (part b)

Every ring is coaxial with the same axis, so the radial parts cancel and the $z$ -parts simply add:

B_{z} = \frac{μ _{0} ρ _{S} ω}{2} \int_{0}^{b} \frac{ρ ^{' 3} d ρ ^{'}}{( ρ ^{' 2} + h ^{2} ) ^{3/2}} .

Substitute $u = ρ^{' 2} + h^{2}$ , so that $ρ^{'} d ρ^{'} = \frac{1}{2} d u$ and $ρ^{'2} = u - h^{2}$ :

\int_{0}^{b} \frac{ρ ^{' 3} d ρ ^{'}}{( ρ ^{' 2} + h ^{2} ) ^{3/2}} = \frac{1}{2} \int_{h^{2}}^{b^{2} + h^{2}} \frac{u - h ^{2}}{u ^{3/2}} d u = [u + \frac{h ^{2}}{u}]_{h^{2}}^{b^{2} + h^{2}} = \frac{b ^{2} + 2 h ^{2}}{b ^{2} + h ^{2}} - 2 h .

Answer

B (0, 0, h) = \frac{μ _{0} ρ _{S} ω}{2} [\frac{b ^{2} + 2 h ^{2}}{b ^{2} + h ^{2}} - 2 h] a_{z} [T]

(b) B_z on the axis, u=h/b. Problem 4.1.3. (a) The spinning disk is a nest of concentric current loops, each carrying dI=\rho_S\omega\rho'\,d\rho', and on the axis their fields simply add. (b) The field is largest at the centre of the disk and has *zero slope* there. It has fallen to half at h\approx0.33\,b and follows the dipole law \mu_0m/2\pi h^3 to better than 5\,\% beyond h\approx4.5\,b. — **Figure 4.1.20.** (b) $B_{z}$ on the axis, $u = h / b$ . Problem 4.1.3. (a) The spinning disk is a nest of concentric current loops, each carrying $d I = ρ_{S} ω ρ^{'} d ρ^{'}$ , and on the axis their fields simply add. (b) The field is largest at the centre of the disk and has *zero slope* there. It has fallen to half at $h \approx 0.33 b$ and follows the dipole law $μ_{0} m /2 π h^{3}$ to better than $5 %$ beyond $h \approx 4.5 b$ .

Check

Units: $μ_{0} ρ_{S} ω b$ is (H/m)(C/m $^{2}$ )(1/s)(m) $=$ (H/m)(A/m) $=$ T. Two limits. $h = 0$ : $B = \frac{1}{2} μ_{0} ρ_{S} ω b a_{z}$ , which is $\int_{0}^{b} μ_{0} d I /2 ρ^{'}$ , the sum of the centre fields of the rings. $h ≫ b$ : expanding the bracket, it tends to $b^{4} /4 h^{3}$ , so $B \to μ_{0} m a_{z} /2 π h^{3}$ with $m = \int_{0}^{b} π ρ^{' 2} d I = π ρ_{S} ω b^{4} /4$ , the field of a magnetic dipole. Far away a spinning charged disk is indistinguishable from a small current loop of that moment (Figure 4.1.20b). For a disk in air $ρ_{S}$ cannot exceed $ε_{0} E_{bd} = 8.842 \times 1 0^{- 12} \times 3 \times 1 0^{6} \approx 27$ μC/m $^{2}$ , and with $b = 10$ cm at $10000$ rpm ( $ω = 1047$ rad/s) this gives $B = \frac{1}{2} μ_{0} ρ_{S} ω b \approx 1.7$ nT at the centre, about $1/30 000$ of the Earth's field. That is why Rowland (1876) needed an astatic magnetometer to prove that convection current is magnetically equivalent to conduction current.

Problem 4.1.3: Rotating charged disk