Loop entering a magnetic field: terminal velocity?
A square conducting loop of side 0.20 m, resistance 0.50 Ω, and mass 0.040 kg falls vertically. Its lower edge crosses a horizontal boundary into a region of uniform magnetic field B=0.80 T pointing horizontally into the page; the loop's plane is in the page. At the instant the lower edge is 0.20 m inside the field, the loop is moving downward at 3.0 m/s and the upper edge is still outside the field. The field region is deep enough that the loop can enter completely before leaving. A student says the loop must quickly reach terminal velocity and then have zero acceleration. Another student says the magnetic force is an internal force and cannot affect the loop's acceleration. The group is asked to resolve the disagreement with a calculation.
1. At the instant described, calculate the induced emf, the current (magnitude and direction), the magnetic force on the loop, and its acceleration.
Read the worked answer
Only the lower horizontal segment is in the field, so the motional emf is ε = B L v = (0.80)(0.20)(3.0) = 0.48 V. The current is I = ε/R = 0.48/0.50 = 0.96 A.
The flux into the page is increasing, so Lenz's law says the induced current must produce a field out of the page. Viewed from the reader's side, that is a counterclockwise current. The lower segment then carries current to the right. The magnetic force on that segment is F = I L B = (0.96)(0.20)(0.80) = 0.154 N, directed upward.
Weight is mg = (0.040)(9.8) = 0.392 N downward. The net force is 0.392 − 0.154 = 0.238 N downward, so a = 0.238/0.040 ≈ 5.95 m/s^2 downward. The acceleration is less than g, and the upward magnetic force is a real external force on the loop, not an internal force.
2. Estimate whether the loop reaches terminal velocity before it is fully inside the field. Calculate the terminal speed and compare with the entry distance and time.
Read the worked answer
Terminal speed occurs when the upward magnetic force equals the weight: mg = B^2 L^2 v_t/R. Thus v_t = mgR/(B^2 L^2) = (0.040)(9.8)(0.50)/[(0.80)^2(0.20)^2] = 0.196/0.0256 ≈ 7.7 m/s.
At 3.0 m/s the loop is far below terminal speed. Even if the acceleration stayed near 6 m/s^2 over the 0.20 m entry distance, v^2 ≈ 3.0^2 + 2(6)(0.20) = 11.4, so v ≈ 3.4 m/s. The entry time is roughly 0.20/3.3 ≈ 0.06 s. The mechanical time constant is τ = m/k, where k = B^2L^2/R = 0.0256/0.50 = 0.051 kg/s, giving τ ≈ 0.040/0.051 ≈ 0.78 s. Since 0.06 s is much less than 0.78 s, the loop does not reach terminal velocity during entry. The first student's 'quickly' claim is false here.
Energetically, at 3.0 m/s the Joule heating rate is I^2R = (0.96)^2(0.50) ≈ 0.46 W, while the gravitational power is mgv ≈ 1.18 W. The magnetic force removes 0.46 W, and the remaining 0.72 W increases kinetic energy. Mechanical energy is not conserved; some becomes Joule heat.
3. What happens if the loop is cut at one point? What happens if it is rotated 90° about a vertical axis while entering?
Read the worked answer
If the loop is cut, no closed conducting path exists, so the induced emf produces no steady current. With I = 0, the magnetic force is zero and the loop falls with acceleration g = 9.8 m/s^2 (ignoring air resistance). No Joule heating occurs, so the loss of gravitational potential energy becomes kinetic energy of the loop rather than electrical energy.
If the loop is rotated 90° about a vertical axis, its plane becomes parallel to the magnetic field, so the magnetic flux through the loop is zero. There is no induced emf and no current while it remains in that orientation. During the rotation itself there is a transient change in flux and a transient current, but afterward the loop falls essentially freely. In both cases the key requirement for electromagnetic induction is a changing magnetic flux through a closed circuit.