Irodov 3.358: The Curl of a Turning Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.358

In some region of an inertial frame the magnetic induction $\vect B$ rotates with angular velocity $\vect{\omega}$ . Find $\nabla\times\vect E$ in this region in terms of the vectors $\vect{\omega}$ and $\vect B$ .

The induction B at a point of the region, at the angle α to the angular velocity ω; its tip runs round the gray circle, counterclockwise seen from the tip of ω. — The induction $\vect B$ at a point of the region, at the angle $\alpha$ to the angular velocity $\vect{\omega}$ ; its tip runs round the gray circle, counterclockwise seen from the tip of $\vect{\omega}$ .

Given

$\vect B$ rotating with angular velocity $\vect{\omega}$

Find

\nabla\times\vect E

in terms of

\vect{\omega}

and

\vect B

Solution

Step 1How fast a rotating vector changes

Let $\alpha$ be the angle between $\vect{\omega}$ and $\vect B$ . In a time $dt$ the vector turns through $\omega\,dt$ about $\vect{\omega}$ , so its tip moves along a circle of radius $B\sin\alpha$ by the arc

|d\vect B| = B\sin\alpha\cdot\omega\,dt

The rate B/ t at the tip of B: tangent to the circle the tip sweeps, perpendicular to both ω and B. — The rate $\partial\vect B/\partial t$ at the tip of $\vect B$ : tangent to the circle the tip sweeps, perpendicular to both $\vect{\omega}$ and $\vect B$ .

That displacement is tangent to the circle, perpendicular to both $\vect{\omega}$ and $\vect B$ , in the sense of the rotation. The vector $\vect{\omega}\times\vect B$ has exactly that direction and the magnitude $\omega B\sin\alpha$ , so at a fixed point of the region

\frac{\partial\vect B}{\partial t} = \vect{\omega}\times\vect B

(1)

Step 2Faraday’s law

The curl of $\vect E$ at each point is minus the local rate of change of $\vect B$ :

\nabla\times\vect E = -\frac{\partial\vect B}{\partial t}

Putting in (1):

\nabla\times\vect E = -\,\vect{\omega}\times\vect B

(2)

Equivalently $\nabla\times\vect E = \vect B\times\vect{\omega}$ : it is perpendicular to both vectors and turns with $\vect B$ , keeping the magnitude $\omega B\sin\alpha$ .

Check

If $\vect B$ lies along the axis, $\alpha = 0$ , the field does not change at all, and (2) gives $\nabla\times\vect E = 0$ , as it must. Units: $\omega B$ is $\text{T/s}$ , and $\nabla\times\vect E$ is $(\text{V/m})/\text{m} = \text{V}/\text{m}^2 = \text{T/s}$ , since a tesla is a $\text{V}\cdot\text{s}/\text{m}^2$ .

Remark

Only the component of $\vect B$ across the axis contributes, and it is the part that turns. This is the situation in the rotating field of a polyphase induction motor: the turning field drives an eddy electric field whose curl turns with it, a quarter turn behind the transverse part of $\vect B$ , and the currents it drives in a conducting rotor drag the rotor round.