Two uniform and antiparallel magnetic fields B1 and B2 exist

Two uniform and anti-parallel magnetic fields, B1 and B2, exist in regions step one another, as shown here. An electron, traveling horizontally in the +x-direction at 966 m/s, enters the B1 field 5.00 mm above the line between B1 and B2. After following an interestingly curvy path, the electron leaves the B1 field 30.0 mm to the right of where it entered, at the same y-value, once again moving horizontally in the +y direction at 966 m/s. If the magnitude of B1 is 1.10 mT. what is the magnitude of B2? How much time from when the electron enters the magnetic fields until it leaves the fields? What is the, least value for B2 for which the electron will leave at that velocity?

Solution

electron q/m = 1.6e-19/9.0e-31 = 1.8e+12

as the electron passes through horizontally, the magnetic force by B1 and B2 will effect its vertical motion only , horizontal velocity in x- direction remains same as 966 m/s

time it takes to pass through

t = 30e-3/966 = 31.05e-6 s

vertical force by B1 = qvB1

vertical acceleration a1 = (q/m)vB1vertically it will travel a distance of 5mm in B1 and moves into B2

time spent in B1 before entering B2 is given by

5e-3 = a1t² = 1.8e+12*966*1.1e-3* t²

t = ±5.12e-8 sec

speed at entry into B2

v_h = 1.8e+12*966*1.1e-3* 5.12e-8 =1.38e+5 m/s

by similarity it will spend the same time after entering B1 from B2

total time spent in B1 = 1.024e-7 sec

time spent in B2 = 31.05e-6 – 1.024e-7 = 30.9 e-6 s

during this time the electron decelerates from v_h to 0 and accelerates back to v_h

time for acceleration/deceleration = 30.9/2 = 15.45e-6 s

acceleration/deceleration in B2 is given by

a2 = (q/m)vB2

15.45e-6 *a2 = v_h = 1.38e+5

a2 = (q/m)vB2 = 1.38e+5/15.45e-6

B2 = 1.38e+5/(1.8e+12*966*15.45e-6) = 5.13 e-6 T

In this situation the electron spends equal time in B1 at the beginning and at the end so that it leaves the field at the same y value

the other option is the electron leaves the region with 0 vertical velocity

at the junction of B1 and B2 with 0 vertical velocity

in this case the electron spends half the time in B1

time spent in B2 = 31.05e-6 – 5.12e-7 = 31.0 e-6 s

time for acceleration/deceleration in B2 = 31.0/2 = 15.5e-6 s

B2 = 1.38e+5/(1.8e+12*966*15.5e-6) = 5.11e-6T

is the next least values of B2 for the electron to leave with same evlocity horizontally