Irodov 3.291: Only the Far End Matters – Amit & Deepali Sinha Foundation

Problem Irodov 3.291

A thin wire $AC$ bent into a semicircle of diameter $d = 20$ cm turns with a constant angular velocity $\vect\omega$ , of magnitude $\omega = 100$ rad/s, in a uniform magnetic field of induction $B = 5.0$ mT, with $\vect\omega$ pointing the same way as $\vect B$ . The axis of rotation passes through the end $A$ and is perpendicular to the diameter $AC$ . Find the line integral of the electric field, $\int \vect E\cdot d\vect r$ , taken along the wire from $A$ to $C$ , and generalize the result.

The semicircular wire AC of diameter d, seen along the rotation axis through A, turning at ω, and the field B parallel to ω. Their common sense is not given; both drawn out of the page. The wire is drawn normal to the axis; the result does not depend on its plane. — The semicircular wire $AC$ of diameter $d$ , seen along the rotation axis through $A$ , turning at $\omega$ , and the field $\vect B$ parallel to $\vect\omega$ . Their common sense is not given; both drawn out of the page. The wire is drawn normal to the axis; the result does not depend on its plane.

Given

$d = 20$ cm
$\omega = 100$ rad/s
$B = 5.0$ mT
$\vect\omega$ parallel to $\vect B$
axis through $A$ , normal to $AC$

Find

\int_A^C \vect E\cdot d\vect r

along the wire; its generalization

Solution

Step 1Inside the wire the electric force cancels the magnetic one

The wire is open, so in the steady state no current flows along it: the free charges in the wire are carried round with it and feel no net force along the wire. The force per unit charge is $\vect E + \vect v\times\vect B$ , and the inertial correction (the centripetal force on the carriers) is smaller than the magnetic term by the factor $m\omega/(eB) \sim 10^{-7}$ , so along every element $d\vect r$ of the wire

\left(\vect E + \vect v\times\vect B\right)\cdot d\vect r = 0 \tag{1}

Integrating along the wire from $A$ to $C$ :

\int_A^C \vect E\cdot d\vect r = -\int_A^C \left(\vect v\times\vect B\right)\cdot d\vect r \tag{2}

An element of the wire at P, its position r from the axis point A, and its velocity v = ω· r, normal to r. The dashed chord AC gives the same integral as the arc. — An element of the wire at $P$ , its position $\vect r$ from the axis point $A$ , and its velocity $\vect v = \vect\omega\times\vect r$ , normal to $\vect r$ . The dashed chord $AC$ gives the same integral as the arc.

Step 2The magnetic force is radial from the axis

Measure $\vect r$ from the point $A$ on the axis, and split it into a part $z$ along the axis and a part $\vect\rho$ perpendicular to it. The element moves at $\vect v = \vect\omega\times\vect r$ . Expanding the triple product with $\vect\omega$ parallel to $\vect B$ :

\begin{aligned} \vect v\times\vect B &= \left(\vect\omega\times\vect r\right)\times\vect B = \vect r\left(\vect\omega\cdot\vect B\right) - \vect\omega\left(\vect r\cdot\vect B\right) \\ &= \omega B\,\vect r - \omega B z\,\hat{\vect z} = \omega B\,\vect\rho \end{aligned}

So the magnetic force per unit charge points straight away from the axis, with magnitude $\omega B\rho$ . Its component along any step of the wire is an exact differential:

\left(\vect v\times\vect B\right)\cdot d\vect r = \omega B\,\vect\rho\cdot d\vect\rho = d\!\left(\frac{\omega B\rho^2}{2}\right) \tag{3}

Step 3The integral along the semicircle

By (3) the integral in (2) depends only on the end points. $A$ is on the axis, $\rho_A = 0$ , and $C$ is a diameter away from it, $\rho_C = d$ :

\int_A^C \vect E\cdot d\vect r = -\left[\frac{\omega B\rho^2}{2}\right]_{0}^{d}

Numerically,

\int_A^C \vect E\cdot d\vect r = -\frac{100\times5.0\times10^{-3}\times0.20^2}{2}\ \text{V}

\int_A^C \vect E\cdot d\vect r = -\frac{\omega Bd^2}{2} = -10\ \text{mV}

(4)

The sign means that $C$ is 10 mV higher in potential than $A$ : with $\vect\omega$ along $\vect B$ the magnetic force pushes positive charge away from the axis, toward $C$ .

Step 4The generalization

Nothing in (3) used the semicircle. For any wire, of any shape, turning rigidly about an axis parallel to a uniform $\vect B$ , with its ends at distances $\rho_1$ and $\rho_2$ from the axis:

\int_1^2 \vect E\cdot d\vect r = -\frac{\omega B}{2}\left(\rho_2^2 - \rho_1^2\right)

(5)

The straight chord $AC$ , a bent wire, or a radial rod of length $d$ from the axis all give (4).

Check

Dimensions: $\omega Bd^2$ is $\text{s}^{-1}\cdot\text{T}\cdot\text{m}^2 = \text{Wb/s} = \text{V}$ . A straight rod of length $a$ turning about one end is the radius of a Faraday disc, and (5) with $\rho_1 = 0$ , $\rho_2 = a$ gives its familiar voltage $\omega Ba^2/2$ in magnitude. Two points at the same distance from the axis, $\rho_1 = \rho_2$ , are at the same potential, whatever wire joins them.

Remark

On charges carried round with the wire, the magnetic force behaves like the force of a potential energy, $-\omega B\rho^2/2$ per unit charge, so its work round any closed wire turning with them is zero. That is why a closed loop spun rigidly about an axis parallel to $\vect B$ has no emf: its flux never changes. Only a sliding contact, as in a Faraday disc with brushes at the center and the rim, lets the radial voltage drive a current.