Question

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 = 0s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.
Required:
Find an expression for the loop's velocity as a function of time as it enters the magnetic field.

Answer 1

Answer:

v₀(1 + B²L²t/mR)

Explanation:

We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Since F = ma where a = acceleration of loop and m = mass of loop

a = F/m = B²L²v₀/mR

Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0

Substituting the value of a and u into v, we have

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)