Answer :

Given:


Speed of electron, v = 5.20 × 106 ms-1


Magnetic field strength normal to the beam = 1.30 × 10–4 T


Charge to mass ratio (e/m) = 1.76 × 1011 C Kg-1


(a) Force applied by magnetic field,


F = e|v × B|


F = e × v × B × sinθ …(1)


Where,


e = Charge on electron


v = velocity of particle


B = magnetic field strength


θ = angle between Magnetic field and velocity


Since the electron traces a circular path, we can use the following equation of centrifugal force:


F = …(2)


Where,


m = mass of particle


v = velocity of particle


r = radius of circle traced


By equating (1) and (2), we get,


evB × sin90° =


r =


r = …(3)


r =


r = 22.7 6 cm


(b) Energy of the electron beam, E = 20 MeV


E = 20 × 106 × 1.6 × 10-19J


The energy of an electron is given by:


E =


From the above equation, we can write,


v = …(4)


putting the value in equation (4)


v =


v = 2.6 × 109 ms-1


This result is incorrect because the speed of any massive object can’t exceed the speed of light (i.e. 3 × 108 ms-1). We can’t use equation (4) in the case where the speed is relativistic:


At relativistic speeds, mass is given by,


m =


Where,


m = relativistic mass


m0 = rest mass


v = velocity of particle


c = speed of light


the radius of the path traced is given by,


r = …(3)


By substituting the value of relativistic mass in equation (3) we get,


r = …(4)


By using equation (4) we can find the radius traced by electrons moving at relativistic speed.


By substituting value of relativistic mass in equation


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