In: Physics
(a) calculate the magnetic field, B required to produce a deflection of 0.25 rad in Thomson’s experiment, assuming the values V= 205 V, length, L = 4.9 cm, and d= 1.5 cm. (b) Find the horizontal velocity vx for this case.
Datas Given
Deflection = 0.25 rad = 0.25 x 180/ =
Potential Difference V = 205 V
Length = 4.9 cm = 0.049 m
d = 1.5 cm = 0.015 m
Electric Field E = V/d = 205 / 0.015 = 13666.67 N/C
a) To find the Magnetic Field B = ?
When the particle aquires avelocity v, then the kinetic energy of the particle = 1/2mv2
ie eV = 1/2mv2
v2 = 2 e V /m = 2 x 1.6 x 10-19 x 205 / 9.1 x 10-31
= 72.08 x 1012
v = = 8.49 x 106 m/s
Now qE = qvBsin
B = E / vsin = 13666.67 / 8.49 x 106 x sin 14.33o
B = 13666.67 / 8.49 x 106 x 0.2470
B = 6517 x 10-6 T
b ) The x- component of velocity vx = v cos = 8.49 x 106 x cos 14.33o
vx = 8.49 x 106 x 0.9690 =8.23 x 106 m/s