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 t

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.

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Gerda 4 years 2021-08-11T07:31:54+00:00 1 Answers 19 views 0

Answers ( )

  1. Adela
    0
    2021-08-11T07:33:19+00:00
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    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)

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