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A thin flat insulating disc of radius 7.5 cm and mass 60 g carries a charge of +4.0 µC, spread…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A thin flat insulating disc of radius $7.5\ \text{cm}$ and mass $60\ \text{g}$ carries a charge of $+4.0\ \mu\text{C}$ , spread evenly over its face. It spins about its own axis at $1.20 \times 10^{3}\ \text{rad/s}$ . The disc is mounted at its centre on a light frictionless gimbal, so gravity gives no torque about the centre and the axis can point in any direction. A uniform magnetic field of $0.80\ \text{T}$ is switched on, and the axis is set at $30^\circ$ to it, as in the figure. The spin is so fast that the angular momentum of the disc is almost all from the spin. Take $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$ .

A thin disc of radius 7.5 cm seen at a slant, mounted at its centre on a pivot. Its axis is tilted 30 degrees from vertical arrows that show a uniform magnetic field of 0.80 T. A curved arrow round the axis shows the disc spinning at 1.20 times 10 to the 3 radians per second. A note gives its mass as 60 g and its charge as plus 4.0 microcoulombs.

Before the field is switched on, what is the magnetic field that the spinning disc makes at its own centre?
What is the magnetic moment of the spinning disc?
What is its angular momentum?
In the field, what is the size of the torque on the disc?
The torque makes the axis turn slowly round the direction of the field, keeping its $30^\circ$ angle. What is the angular speed of this slow turning?
How long does the axis take to go once round the field?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. A spinning charged disc is a set of current rings. Add their fields and moments. The torque in the field then acts on the disc like gravity on a spinning top: the axis turns slowly round the field.

Step 1. Field at the centre (a). With $\sigma = \dfrac{Q}{\pi R^2}$ , a ring carries $dI = \sigma\omega r\,dr$ , so

\displaystyle B = \int_0^R \frac{\mu_0\sigma\omega r\,dr}{2r} = \frac{\mu_0\sigma\omega R}{2} = \frac{\mu_0 Q\omega}{2\pi R} = 12.8\ \text{nT}

Step 2. Moment (b). $m = \displaystyle\int_0^R \pi r^2\sigma\omega r\,dr = \frac{\pi\sigma\omega R^4}{4} = \frac{Q\omega R^2}{4} = 6.75 \times 10^{-6}\ \text{A m}^2$ .

Step 3. Angular momentum (c). $L = \tfrac12 MR^2\omega = 0.2025\ \text{kg m}^2/\text{s}$ . Note that $\dfrac{m}{L} = \dfrac{Q}{2M}$ : the same for every ring, because charge and mass are spread in the same way.

Step 4. Torque (d). $\tau = mB_0\sin30^\circ = 2.70 \times 10^{-6}\ \text{N m}$ , at right angles to both $\vec m$ and the field.

Step 5. The slow turning (e). The torque is at right angles to $\vec L$ , so it changes only its direction. The tip of $\vec L$ moves on a circle of radius $L\sin\theta$ , so $\tau = \Omega L\sin\theta$ and

\displaystyle \Omega = \frac{mB_0}{L} = \frac{QB_0}{2M} = 2.67 \times 10^{-5}\ \text{rad/s}

It does not depend on the angle or on the spin.

Step 6. One turn (f). $T = \dfrac{2\pi}{\Omega} = 2.36 \times 10^{5}\ \text{s} = 65.4\ \text{h}$ .

Answer: (a) $12.8\ \text{nT}$ . (b) $6.75 \times 10^{-6}\ \text{A m}^2$ . (c) $0.2025\ \text{kg m}^2/\text{s}$ . (d) $2.70 \times 10^{-6}\ \text{N m}$ . (e) $2.67 \times 10^{-5}\ \text{rad/s}$ . (f) $65.4\ \text{h}$ .

Watch out. The torque does not swing the axis into line with the field, as it would for a compass needle that is not spinning: it turns the axis round the field. And putting all the charge at the rim gives a moment of $Q\omega R^2/2 = 1.35 \times 10^{-5}\ \text{A m}^2$ , twice too large.

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