Electromagnetic induction > Vortex electric field > Problem 11.2.3

Statement

$11.2.3.$

A conducting ring having a jumper with an electric light bulb in diameter is
moved in the magnetic field of the solenoid with current so that the plane of
the ring is perpendicular to the axis of the solenoid, and the jumper with the
light bulb is perpendicular to the direction of the speed of movement of the
ring. In ring positions A and B, the light bulb glows, and in position C, it goes
out. Explain the observed effect.

Solution

A long solenoid with current produces a magnetic field with the following characteristics:

· Inside, far from the ends, the field is uniform and parallel to the axis (position C).
· Near the ends (positions A and B), the field is non‑uniform: the field lines spread out and there is both a radial and an axial gradient. The intensity decreases as one moves away from the center.

Taking this into account, let us analyze the motion of the ring and the magnetic flux.

The ring moves horizontally, keeping its plane perpendicular to the axis of the solenoid.

· At A and B (ends): when the ring moves, it passes through regions where the field intensity varies with position. Therefore, the magnetic flux through the area of the ring changes with time.

At C (center): the field is uniform. When shifting the ring laterally, the value of$ \mathbf{B}$ at every point on its surface is the same, and the flux then remains constant.

We apply Faraday's Law.

The electromotive force induced in the closed circuit (ring + bridge) is:

$\mathcal{E} = -\frac{d\Phi}{dt}$

· At A and B
$ d\Phi/dt \neq 0 $ → an emf is induced → current flows through the bulb → it lights up.
· At C
$ d\Phi/dt = 0$ → there is no emf → no current flows → the bulb turns off.

Answer

$\boxed{d\Phi/dt \neq 0}$

$\boxed{ d\Phi/dt = 0}$