Irodov 3.290: Inertia Against the Lorentz Force – Amit & Deepali Sinha Foundation

Problem Irodov 3.290

A metal disc of radius $a = 25$ cm spins about its own axis with a constant angular velocity $\omega = 130$ rad/s. Find the potential difference between its center and its rim

when there is no external magnetic field;
when a uniform external magnetic field of induction $B = 5.0$ mT is applied perpendicular to the disc.

Given

$a = 25$ cm
$\omega = 130$ rad/s
(b) $B = 5.0$ mT normal to the disc
$m = 9.11\times10^{-31}$ kg
$e = 1.60\times10^{-19}$ C

Find

\Delta\varphi

between center and rim, (a) and (b)

Solution

Step 1What holds an electron on its circle

A free electron at distance $r$ from the axis is carried round a circle at speed $v = \omega r$ , so the net force on it must be the centripetal force $m\omega^2 r$ toward the axis. In the steady state the conduction electrons have redistributed until the fields inside the metal supply exactly that force. Writing $\hat{\vect r}$ for the outward radial unit vector and $-e$ for the electron’s charge:

-e\left(\vect E + \vect v\times\vect B\right) = -m\omega^2 r\,\hat{\vect r} \tag{1}

Step 2(a) No field: the electric field alone

With $\vect B = 0$ , (1) fixes the electric field inside the disc. It points radially outward, so that its force on the electron, $-e\vect E$ , points inward:

\vect E = \frac{m\omega^2 r}{e}\,\hat{\vect r}

The potential falls along $\vect E$ , so the center is the higher. Integrating from the center to the rim:

\varphi_{\mathrm{center}} - \varphi_{\mathrm{rim}} = \int_0^a E\,dr = \frac{m\omega^2}{e}\int_0^a r\,dr = \frac{m\omega^2a^2}{2e}

Numerically,

\Delta\varphi = \frac{9.11\times10^{-31}\times130^2\times0.25^2}{2\times1.60\times10^{-19}}\ \text{V}

\Delta\varphi = \frac{m\omega^2a^2}{2e} = 3.0\ \text{nV}

(2)

The center is positive with respect to the rim: the electrons have been flung outward until the field holds them in.

Part (b): an element dr on a radius, moving at v normal to it, and the polarity of center and rim set by v· B for the senses drawn. — Part (b): an element $dr$ on a radius, moving at $\vect v$ normal to it, and the polarity of center and rim set by $\vect v\times\vect B$ for the senses drawn.

Step 3(b) With the field: the magnetic force dominates

Now $\vect v = \vect\omega\times\vect r$ is normal to the radius, and with $\vect B$ normal to the disc the magnetic force $-e\,\vect v\times\vect B$ is radial, of magnitude $e\omega rB$ . Its ratio to the centripetal term of (1) is

\frac{m\omega^2 r}{e\omega rB} = \frac{m\omega}{eB} = \frac{9.11\times10^{-31}\times130}{1.60\times10^{-19}\times5.0\times10^{-3}} \approx 1.5\times10^{-7}

so the centripetal term is negligible, and (1) reduces to the electric field cancelling the magnetic force:

\vect E \approx -\,\vect v\times\vect B, \qquad E = \omega rB

Integrating along a radius from the center to the rim:

|\Delta\varphi| = \int_0^a \omega Br\,dr

Numerically,

|\Delta\varphi| = \frac{130\times5.0\times10^{-3}\times0.25^2}{2}\ \text{V}

|\Delta\varphi| \approx \frac{\omega Ba^2}{2} = 20\ \text{mV}

(3)

For the senses drawn, counter-clockwise rotation with $\vect B$ out of the page, $\vect v\times\vect B$ points outward and the rim is positive; reversing either sense makes the center positive.

Check

Dimensions: $m\omega^2a^2/e$ is $(\text{kg}\cdot\text{m}^2/\text{s}^2)/\text{C} = \text{J/C} = \text{V}$ , and $\omega Ba^2$ is $\text{s}^{-1}\cdot\text{T}\cdot\text{m}^2 = \text{Wb/s} = \text{V}$ . The ratio of the two answers is the same small number found above, as it must be:

\frac{m\omega^2a^2/2e}{\omega Ba^2/2} = \frac{m\omega}{eB} \approx 1.5\times10^{-7}

Remark

Part (b) is the Faraday disc. Its emf is also the flux cut per unit time by one radius: in one turn, which takes $2\pi/\omega$ , a radius sweeps the whole area $\pi a^2$ , so

\frac{B\pi a^2}{2\pi/\omega} = \frac{\omega Ba^2}{2}

The inertial voltage of part (a) is real but tiny; the Tolman and Stewart experiments detected its transient cousin, the pulse of current when a spinning coil is stopped, and so measured $e/m$ for the carriers in a metal.