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.]
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
= 90o
L = 5 mm
= 0.005 m
arc length L = 2
r (
/ 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.
5) An electron gun sends out a beam of electrons whose kinetic energies are all about...
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that causes it to turn through a 90∘∘ circular arc of length 3 mm.
How strong must the magnetic field be? [Note: electrons have a mass
of about 9.11×10-31 kg.]
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