If a charged particle is moving in a uniform magnetic field, you found that it w

Question

If a charged particle is moving in a uniform magnetic field, you found that it will move in circular motion - provided that the speed remains constant (note that you found that the radius of the circular motion is proportional to the speed). Since the magnetic force is always perpendicular to the motion, the magnetic force doesn't do any work on the particle -so it does not change its speed. So if we only have a uniform magnetic field, the charged particle will move in circles of constant radius. An electron is moving in a 3.9 mT (milliTesla) magnetic field with angular momentum of 6.4 Times 10^-26 kg m^2/s, perpendicular to the magnetic field. What is the radius of the electron's circular motion? What is the frequency of its circular motion?

Explanation / Answer

in the circular motion, centripetal force balances the magnetic force.

hence q*v*B=m*v^2/r

where q=charge on the electron

v=speed of electron

B=magnetic field

m=mass of electron

r=radius of the path

hence r=m*v/(q*B)=9.1*10^(-31)*v/(1.6*10^(-19)*3.9*10^(-3))=1.4583*10^(-9)*v...(1)

its angular momentum is given by m*v*r.

hence m*v*r=6.4*10^(-26)

==>9.1*10^(-31)*1.4583*10^(-9)*v^2=6.4*10^(-26)

==>v=sqrt(6.4*10^(-26)/(9.1*10^(-31)*1.4583*10^(-9)))=6.9446*10^6 m/s

then r=1.4583*10^(-9)*v=0.0101 m

frequency of circular motion=1/time period

=1/(circumference/speed)

=speed/circumference

=v/(2*pi*r)

=(6.9446*10^6)/(2*pi*0.0101)=1.0943*10^8 Hz

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