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

Problem 5.3.2

Interactive: Faraday's Law of Induction

Problem 5.3.2

The Faraday disk generator

The Faraday disk generator consists of a circular metal disk rotating with a constant angular velocity $ω$ in a uniform and constant magnetic field of $B = B_{0} a_{z}$ that is parallel to the axis of rotation. Brush contacts are provided at the axis and on the rim of the disk, as shown in the figure. Determine the open-circuit voltage of the generator if the radius of the disk is $b$ .

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 7-3.

Try it yourself first.

Strategy

$B$ is constant, so the transformer term of Eq. (5.3.10) vanishes and $e = \oint_{C} (v \times B) \cdot d l$ . The only moving part of the circuit $1 \to 2 \to 2^{'} \to 3 \to 4 \to 1^{'} \to 1$ is the radial path $3 \to 4$ inside the disk. The external wires and the brushes are at rest and contribute nothing.

1. Velocity and

v \times B

A point of the disk at radius $ρ$ moves on a circle with the speed $ω ρ$ , so $v = ω ρ a_{ϕ}$ and (Figure 5.3.4a)

v \times B = ω ρ a_{ϕ} \times B_{0} a_{z} = ω ρ B_{0} a_{ρ},

a force per unit charge directed radially outward, growing linearly with $ρ$ .

2. The line integral

Go round the circuit in the sense $1 \to 2 \to 2^{'} \to 3 \to 4 \to 1^{'} \to 1$ . On the segment $3 \to 4$ , from the rim to the centre, $d l = d ρ a_{ρ}$ with $ρ$ running from $b$ to $0$ :

V_{0} = \oint_{C} (v \times B) \cdot d l = \int_{b}^{0} ω B_{0} ρ d ρ = - \frac{1}{2} ω B_{0} b^{2} .

Answer

V_{0} = - \frac{1}{2} ω B_{0} b^{2} [V] (∣ V_{0} ∣ = \frac{1}{2} ω B_{0} b^{2})

3. Reading the sign

With terminal $1$ connected to the axis, as drawn, the meter reads negative: $v \times B$ pushes positive carriers outward, so the rim is the positive terminal. Reverse $ω$ or $B_{0}$ and the sign reverses.

Vary $B_{0}$ , $b$ and the speed and check $∣ V_{0} ∣ = 0.785$ V at the book's values.

InteractiveLoading simulation…

(a) Top view: \vec v\times\vec B is radial. — (a) Top view: $v \times B$ is radial.

(c) |V_0| against speed, B_0=0.5\,\mathrm{T}. Problem 5.3.2. (a) Every point of the disk moves along \vec a_\phi, and \vec v\times\vec B=\omega\rho B_0\vec a_\rho pushes positive carriers towards the rim. (b) For B_0=0.50\,\mathrm{T} and 3000\,\mathrm{rpm} the voltage grows as \rho^2 along a radius and reaches 0.785\,\mathrm{V} at b=100\,\mathrm{mm}: half of it is generated in the outer 29\,\% of the radius. (c) |V_0|=\frac12\omega B_0b^2 is proportional to the speed and to the square of the radius. — **Figure 5.3.4.** (c) $∣ V_{0} ∣$ against speed, $B_{0} = 0.5 T$ . Problem 5.3.2. (a) Every point of the disk moves along $a_{ϕ}$ , and $v \times B = ω ρ B_{0} a_{ρ}$ pushes positive carriers towards the rim. (b) For $B_{0} = 0.50 T$ and $3000 rpm$ the voltage grows as $ρ^{2}$ along a radius and reaches $0.785 V$ at $b = 100 mm$ : half of it is generated in the outer $29 %$ of the radius. (c) $∣ V_{0} ∣ = \frac{1}{2} ω B_{0} b^{2}$ is proportional to the speed and to the square of the radius.

Check with numbers

For $B_{0} = 0.50 T$ , $b = 100 mm$ and $3000 rpm$ , $ω = 2 π \times 50 = 314.2 rad/s$ and

∣ V_{0} ∣ = \frac{1}{2} \times 314.2 \times 0.50 \times (0.1)^{2} = 0.785 V,

while the field inside the disk reaches $E = ω B_{0} b = 314.2 \times 0.5 \times 0.1 = 15.7 V/m$ at the rim (Figure 5.3.4b). Units: $ω B_{0} b^{2}$ is $s^{- 1} \cdot T \cdot m^{2} = Wb/s = V$ . Nothing about the circuit changes in time, its shape is fixed and $B$ is constant, so there is no obvious $d Φ/ d t$ : this homopolar machine is the clearest case in which one must return to the motional integral $\oint (v \times B) \cdot d l$ instead of the flux rule. Because $V_{0} \propto ω b^{2}$ while the rim speed $ω b$ is limited mechanically to a few hundred m/s (for $100 V$ with $B_{0} = 1 T$ and $b = 0.5 m$ one would need $ω = 800 rad/s$ , a rim speed of $400 m/s$ ), such machines are built the other way round: very low voltage and very large current, with the brushes as the hard part.

Problem 5.3.2: The Faraday disk generator