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Question:

In the x-y-plane, the region y>0 has a uniform magnetic field B1^k and the region y<0 has another uniform magnetic field B2^k. A positively charged particle is projected from the origin along the positive y-axis with speed Vo=π m/s at t=0, as shown in the figure. Neglect gravity in this problem. Let t=T be the time when the particle crosses the x-axis from below for the first time. If B2=4B1, the average speed of the particle, in m/s, along the x-axis in the time interval T is ________.

Solution:

Average speed along the x-axis (Vx) = ∫|Vx|dt / ∫dt = (d1+d2)/(t1+t2) →(1)
We also have, r1=mv/qB1, r2=mv/qB2
since B1 = B2/4 ∴ r1=4r2 →(2)
Time in B1 → π/qB1 = t1
Time in B2 → π/qB2 = t2
Total distance along x-axis d1+d2 = 2r1+2r2 = 2(r1+r2) = 2(5r2)
Total time T = t1+t2 = 5t2
∴ Average speed = 10r2/5t2 = 2mv/qB2 × qB2/πm = 2