Question

5) An electron gun sends out a beam of electrons whose kinetic energies are all about 66 μμeV. (1 μeV=1.6×10−25 J1 μeV=1.6×10−25 J.) You need to set up a magnetic field perpendicular to the beam that causes it to turn through a 90∘∘ circular arc of length 5 mm. How strong must the magnetic field be? [Note: electrons have a mass of about 9.11×10^-31 kg.]

Answer 1

given

mass of electron m = 9.11 x 10^-31 kg

electron charge is q = 1.6 x 10^-19 C

KE = 66 μeV

= 66 x 1.6 x 10^-25 J

= 1.056 x 10^-23 J

angle $\theta$ = 90^o

L = 5 mm

= 0.005 m

arc length L = 2 $\pi$ r ( $\theta$ / 360 )

0.005 = 2 x 3.14 x r x ( 90 / 360 )

0.005 = 2 x 3.14 x r x ( 1/4 )

r = 0.005 x 4 / 2 x 3.14

r = 3.184 x 10^-3 m

KE = 1/2 m v²

1.056 x 10^-23 = 0.5 x 9.11 x 10^-31 x v²

v = 4814.9 m/sec

using equation

F = m v² / r and F = q v B sin90 ( given magnetic field perpendicular to the beam )

m v² / r = q v B sin90

m v² / r = q v B

m v / r = q B

r = m v / q B

3.184 x 10^-3 = 9.11 x 10^-31 x 4814.9 / 1.6 x 10^-19 x B  

B = 9.11 x 10^-31 x 4814.9 / 1.6 x 10^-19 x 3.184 x 10^-3  

B = 8.61 x 10^-6 T

or

B = 8.61 μ T

so the magnetic field is B = 8.61 x 10^-6 T or 8.61 μ T strong.