Problem — Loop entering a field (qualitative)

Problem

A rectangular loop is pushed at constant velocity $v$ into a region where there is a uniform field $B$, perpendicular to the plane of the loop and pointing into the page. (a) During entry (when only one side is already inside the field), in which direction does the induced current flow? In which direction does the magnetic force act on the loop? (b) When the loop is fully immersed in the field, what is the current $i$? (c) During exit, in which direction does the induced current flow? (d) To keep $v$ constant one must push: where does the work done end up?

Solution

(a) Entry. As the loop enters, the area immersed in the field grows, so the inward flux through the loop increases. By Lenz’s law the induced current opposes this increase: it must generate a field pointing out inside the loop, so it flows counterclockwise (with $B$ into the page). On the leading immersed side, carrying this current in the presence of $B$, the Laplace force $\vv{F}=i\,\vv{\ell}\times\vv{B}$ acts: it turns out to be directed opposite to $\vv{v}$, i.e. it brakes the entry.

Note: the exact direction of the current depends on the orientation of $\vv{B}$; the principle (Lenz: oppose the change in flux) remains valid in every case.

(b) Loop fully inside. With the loop entirely immersed, the area within the field no longer changes: the flux $\Phi_B=B\cdot A$ is constant. $\text{emf}=-\ddt{\Phi_B}=0\quad\Longrightarrow\quad i=0$ There is no current and no magnetic force: the loop would slide without opposition.

(c) Exit. Now the immersed area decreases, the inward flux drops. By Lenz’s law the induced current opposes the decrease: it flows in the opposite direction from entry (clockwise), to reinforce the vanishing inward field. In this phase too the Laplace force opposes the motion, braking the exit.

(d) Energy balance. During entry and exit the magnetic force brakes the loop; to keep $v$ constant the external agent must do work against it. This work does not increase the kinetic energy (the velocity is constant): it is entirely dissipated in the loop’s resistance $R$ by the Joule effect, $P=Ri^2$, heating the conductor.

$\ev{\text{(a) counterclockwise, force opposite to }\vv{v};\quad \text{(b) } i=0;\quad \text{(c) opposite direction;}\quad \text{(d) Joule heat in } R}$

Links

Topics: Electromagnetic induction Concepts: Lenz’s law · Faraday-Neumann-Lenz law · Laplace force · Joule effect