How to Calculate the Induced EMF of a Spinning Metal Rod

A spinning conducting rod can develop a voltage between its ends because its charges move through a magnetic field. The result depends on how the rod is mounted and how the field is oriented, so first identify the geometry.

For the standard arrangement—a rod rotating about one end in a plane, with a uniform magnetic field perpendicular to that plane—the emf between pivot and tip has magnitude $\frac{1}{2}B\omega L^2$. The derivation also shows how to determine its sign and how the answer changes if the rod rotates about its center.

Motional emf along a moving rod

A charge $q$ moving with the rod at velocity $\mathbf{v}$ experiences the magnetic Lorentz force $q(\mathbf{v}\times\mathbf{B})$. The magnetic contribution to the emf along a conductor from point $a$ to point $b$ is

$$\mathcal{E}_{a\to b}=\int_a^b(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}.$$

This line integral accounts for the fact that different points on a spinning rod have different velocities. In an open-circuit rod, charges redistribute until the electric force balances the magnetic force. Under the usual idealized conditions, the resulting potential difference between the endpoints has the magnitude given by this integral.

Rod rotating about one end

Let the rod have length $L$ and rotate at constant angular speed $\omega$ about a pivot at one end. Let the magnetic field be uniform and perpendicular to the plane of rotation. At a distance $r$ from the pivot, the rod element moves at speed

$$v(r)=\omega r.$$

The velocity is perpendicular to the rod, and with the field normal to the plane, $\mathbf{v}\times\mathbf{B}$ points along the rod. A small element of length $dr$ therefore contributes an emf of magnitude $B\omega r\,dr$. Integrating from the pivot to the tip gives

$$\begin{aligned} |\mathcal{E}|&=\int_0^L B\omega r\,dr\\ &=B\omega\left[\frac{r^2}{2}\right]_0^L\\ &=\frac{1}{2}B\omega L^2. \end{aligned}$$

The emf is not simply $B\omega L^2$: the inner parts of the rod move more slowly than the tip, so the speed must be integrated over its length.

If the magnetic field is not perpendicular to the plane of rotation, replace $B$ with its signed perpendicular component $B_\perp=\mathbf{B}\cdot\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is a unit vector normal to that plane. A field component lying in the plane does not contribute to the integral along this radial rod.

Worked example

A $0.50\,\text{m}$ rod rotates at $12\,\text{rad/s}$ in a $0.40\,\text{T}$ field perpendicular to its plane. Its pivot is at one end. The emf magnitude is

$$\begin{aligned} |\mathcal{E}|&=\frac{1}{2}(0.40)(12)(0.50)^2\\ &=0.60\,\text{V}. \end{aligned}$$

Thus the tip and pivot differ in potential by $0.60\,\text{V}$ in the ideal open-circuit model. To identify which is at higher potential, specify the rotation direction and magnetic-field direction. For example, with the rod pointing radially outward, rotation counterclockwise as viewed from above, and the field pointing upward, $\mathbf{v}\times\mathbf{B}$ points outward. Positive charges are driven toward the tip, making it the higher-potential end.

If the rod rotates about its center

A different result applies when the pivot is at the rod’s midpoint. Write the position along the rod as $s$, measured from the center, so the two ends are at $s=\pm L/2$. For the same perpendicular field and rotation, the emf from the center to a point at position $s$ is proportional to

$$\int_0^s B\omega s'\,ds'=\frac{1}{2}B\omega s^2.$$

Each endpoint is therefore separated from the center by an emf of magnitude $B\omega L^2/8$. But the two ends have the same potential in this symmetric setup: the contributions from the two halves cancel when integrating from one end to the other. So the end-to-end emf is zero, even though there is a nonzero center-to-end voltage.

Common mistakes

• Using the tip speed for the whole rod. The speed varies as $\omega r$, which is why the integration produces the factor $1/2$.
• Ignoring the field direction. The relevant field is its component normal to the plane of rotation; reversing that component reverses the polarity.
• Confusing rotation about an end with rotation about the center. The first gives a nonzero pivot-to-tip voltage; the second gives equal potentials at the two ends in the symmetric case.
• Treating the rod as a complete circuit. An emf along a rod is a voltage between points in a specified measurement arrangement. If connecting wires or a closed circuit also move through the field, their motional-emf contributions may need to be included.

The general method is to calculate the velocity at each point, form $\mathbf{v}\times\mathbf{B}$, and integrate its component along the conductor. This works beyond the simple perpendicular-field example whenever the geometry is specified.