An electron at rest of mass 9.11x10^−31 kg is accelerated through a potential difference of 350 V. It then enters some deflecting plates of 50 V with dimensions as shown. Calculate the distance, or of the deflection of the electron. The charge on an electron is 1.6x10^−19 C). (20 marks)

r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}r=

B

1

e

2Vm

Explanation:

Let m and e are the mass and charge of an electron. It is accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. Let v is the velocity of the electron. It can be calculated as :

\dfrac{1}{2}mv^2=eV

2

1

mv

2

=eV

v=\sqrt{\dfrac{2eV}{m}}v=

m

2eV

When the electron enters the magnetic field, the centripetal force is balanced by the magnetic force as :

\dfrac{mv^2}{r}=evB

r

mv

2

=evB

r=\dfrac{mv}{eB}r=

eB

mv

or

r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}r=

B

1

e

2Vm

So, the radius of the resulting electron trajectory is \dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}

B

1

e

2Vm

. Hence, this is the required solution.

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