Lets look at the details of eddy-current braking. A square loop, length l on each side, is

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Let’s look at the details of eddy-current braking. A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0 s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0 m.
a. Find an expression for the loop’s velocity as a function of time as it enters the magnetic field. You can ignore gravity, and you can assume that the back edge of the loop has not entered the field.
b. Calculate and draw a graph of v over the interval 0 s ≤ t ≤ 0.04 s for the case that v0 = 10 m/s, l = 10 cm, m = 1.0 g, R = 0.0010 Ω, and B = 0.10 T. The back edge of the loop does not reach the field during this time interval.

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