Question

Figure shows a rectangular loop being pulled out of a magnetic field \( B \) with a velocity \( v \) by applying an external force \( F \). Find the amount of magnetic force induced on the loop and show its direction.

Hint:

Only the vertical arm \( l_1 \) is in the field: find the motional EMF \( Bl_1v \), then the induced current, then \( F=BIl_1 \); direction opposes the motion (Lenz's law).

Answer 1

Step 1: Identify the arm that cuts the field.
The loop (vertical side \( l_1 \), horizontal side \( l_2 \)) is pulled to the right, out of a region where \( B \) points into the page. As it moves, only the left vertical arm of length \( l_1 \) is still inside the field and cuts the magnetic field lines.

Step 2: Motional EMF.
The EMF induced in this arm is
\[ \varepsilon = B\,l_1\,v \]

Step 3: Induced current.
If \( R \) is the total resistance of the loop, the induced current is
\[ I = \frac{\varepsilon}{R} = \frac{B\,l_1\,v}{R} \]

Step 4: Force on the current-carrying arm.
This current, flowing through the arm of length \( l_1 \) that lies in the field, experiences a force
\[ F_m = B\,I\,l_1 = B\left(\frac{B\,l_1\,v}{R}\right)l_1 \]
\[\boxed{F_m = \frac{B^2 l_1^{\,2}\,v}{R}}\]

Step 5: Direction.
By Lenz's law the induced effects oppose the change causing them. Since the flux (into the page) is decreasing as the loop leaves, the induced force acts to oppose the withdrawal. Hence \( F_m \) points to the left, opposite to the applied force \( F \) and to the velocity \( v \). To keep the loop moving at constant speed, the external force must satisfy \( F = F_m \).