What accelerating voltage would be necessary to produce an e

What accelerating voltage would be necessary to produce an electron orbit with diameter 10 cm if the magnetic induction perpendicular to the electron beam is 0.001 Tesla. The book value of e/m is provided at the end of the text. 2. For the boxed \"Lorentz Force Demonstrator\" units, what is the magnetic induction at the center of the Helmholtz coils for currents of 0.9 A: 1.2 A, and 1.5 A. Give the result in the still popular non-SI units Gauss (1 Tesla = 10A4 Gauss) rounded to the first decimal place.

Solution

1)

For circular motion:

the magnetic force must provide the necessary centripetal force:

So, qvB = mv^2/R

Now, the Kinetic energy is provided by the accelerated voltage V

So, 0.5*mv^2 = Vq

So, mv^2 = 2Vq

So, qvB = 2Vq/R

So, V = vBR/2

Now, v = sqrt(2Vq/m)

So, V = sqrt(2Vq/m)*B*R/2

So, V = sqrt(2*V*1.6*10^-19/(9.1*10^-31))*0.001*0.05/2

So, V = 219.8 V <---------answer

2)

Magnetic field at the center is given by:

B = u*I/(2*r)

So, for I = 0.9 A

= 1.26*10^-6*0.9/(2*0.05)

= 1.14*10^-5 T

= 1.14*10^-1 G = 0.114 G <------- in Gauss

Similarly for I = 1.2 A

B = 1.26*10^-6*1.2/(2*0.05)

= 1.51*10^-5 T

= 0.151 G

and for I = 1.5 A

B = 1.26*10^-6*1.5/(2*0.05)

= 1.89*10^-5 T

= 0.189 G