Loop Entering a Field Region at Constant Speed
A rectangular conducting loop of width w = 0.20 m (along the direction of motion) and height h = 0.10 m has total resistance R = 0.50 Ω. It moves at constant speed v = 3.0 m/s into a region of uniform magnetic field B = 0.40 T pointing out of the page, confined to a strip of width 0.50 m. The loop enters at t = 0 and is fully inside the strip from some later time until it begins to exit. An external agent pulls the loop so that its speed stays constant throughout. A student claims the induced emf is zero whenever the loop is moving, since the flux through the loop is constant in magnitude once fully inside.
1. Determine the induced emf and current (magnitude and direction) while the loop is entering the field, and the external force required.
Read the worked answer
While entering, the flux Φ = B·(area inside) = B·h·(vt) increases at rate dΦ/dt = Bhv = (0.40)(0.10)(3.0) = 0.12 V. Induced emf = 0.12 V. Current I = emf/R = 0.12/0.50 = 0.24 A. By Lenz's law the induced current must oppose the increasing outward flux, so it produces an inward field: clockwise as viewed with B out of the page. The magnetic force on the leading side is F = IhB = (0.24)(0.10)(0.40) = 9.6×10⁻³ N, opposing motion; the external agent must supply this force (and power P = Fv = 2.9×10⁻² W, equal to I²R).
2. Describe the emf and current during the fully-inside phase and during exit, and explain the student's error precisely.
Read the worked answer
Fully inside, the flux is constant (Φ = B·h·w = 8.0×10⁻³ Wb), so dΦ/dt = 0 and emf = 0; the student is correct there. During exit, the flux decreases at rate Bhv = 0.12 V again, so emf = 0.12 V but the current reverses (counterclockwise) to oppose the decrease, and the magnetic force again opposes motion. The student's error is generalizing 'moving' to 'changing flux': motion alone does not induce emf in a rigid loop; only the rate of change of enclosed flux does. Note that motional emf Bhv along the leading side exists, but the net loop emf is zero when both vertical sides are in the field because the contributions cancel.