Question

An electron enters a magnetic field of 0.14 T with a velocity perpendicular to the direction of the field. What is the value f ×10-10, where f is the frequency (in Hz) at which the electron traverse a circular path?

Answer #1

Solution:

Mass of an electron = m = 9.1 e-31 kg

Charge on an electron = q =1.6 e-19 C

Magnetic field in which it enters = B = 0.14 T

Since it moves in a circular path there exists centripetal force
=mv^{2}/R, where R = radius of the pathand v its
velocity

and F= qvB due to the magnetic field.

mv^{2}/R = qvB

v = qBR/ m

= (1.6 e-19)(0.14) R / (9.1 e-31)

= 0.0246 R e 12 m/s

Distance traveled in a complete circle = 2 pi R

v = 2 pi R /T =2 pi R f

f = v / 2 pi R = (0.0246 R e 12) / 2 pi R

= 0.0039 e12

= **3.92 e9 Hz**

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