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: Use the rate of change of flux.
Let the length of the loop still inside the field be \(x\) (measured horizontally), while the vertical side in the field has length \(l_1\). The flux linked is \(\phi = B\,l_1\,x\).

Step 2: Induced EMF from Faraday's law.
As the loop is pulled out, \(x\) decreases at rate \(\dfrac{dx}{dt} = -v\). The magnitude of the induced EMF is
\(\varepsilon = \left|\dfrac{d\phi}{dt}\right| = B\,l_1\left|\dfrac{dx}{dt}\right| = B\,l_1 v.\)

Step 3: Induced current and retarding force.
Current \(I = \dfrac{\varepsilon}{R} = \dfrac{B l_1 v}{R}\). The power delivered by the retarding magnetic force equals the electrical power dissipated: \(F_m\,v = I^2 R\).

Step 4: Solve for the force.
\(F_m = \dfrac{I^2 R}{v} = \dfrac{1}{v}\left(\dfrac{B l_1 v}{R}\right)^2 R = \dfrac{B^2 l_1^{\,2} v}{R}.\)
\[\boxed{F_m = \frac{B^2 l_1^{\,2} v}{R}}\]

Step 5: Direction of the force.
Energy conservation and Lenz's law both demand that this magnetic force oppose the motion. It therefore acts leftward, back into the field region, directly opposing the pull \(F\). The applied force does work that appears entirely as heat \(I^2R\) in the loop.