Question

An electron enters a region of space containing a uniform 0.0000149-T magnetic field. Its speed is 121 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron\'s path, and the frequency f of the motion.

Answer 1

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apply centripetal force = magnetic force

i.e mv^2/r = qvB,

where m = mass o the charged particle

v = velocity,   r = radius ,

q = charge = 1.6*10^-19 C

B = magnetic field

so now , r = mv/qB

so r    = 9.11e-31 * 121/(1.6e-19*0.0000149)

r = 0.046 mm

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T = 1/f = 2piR/V

= 2pi (mv)/qBv

= 2pim/qB

T = 2pim/qB

f = qB/2pim

f = 1.6e-19 * 0.0000149 /(2*3.14* 9.11e-31)

f = 4.16 *10^5 Hz
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