An electron is accelerated from rest towards the right in the horizontal directi

Question

An electron is accelerated from rest towards the right in the horizontal direction by a potential difference of 400 V. The electron then enters a uniform magnetic field of magnitude 0.2 T directed into the page with its velocity perpendicular to the field. (a) What are the magnitude and direction of an electric field that will cause the electron to continue to move horizontally? Ignore gravity. The electric field is now removed, (c) Calculate the radius of the electron's path in the magnetic field.

Explanation / Answer

q = charge on electron = 1.6 x 10-19 C

V = potential difference = 400 Volts

m = mass of electron = 9.1 x 10-31 kg

v = speed gained by the electron

Using conservation of energy ::

KE = Electric Potential energy

(0.5) mv2 = qV

inserting the values

(0.5) (9.1 x 10-31 ) v2 = (1.6 x 10-19) (400)

v = 1.186 x 107 m/s        towards right

E = electric field

B = magnetic field = 0.2 T   into the page

a) for electron to pass undeflected

electric force = magnetic force

qE = qvB Sin90

E = vB

E = (1.186 x 107) (0.2)

E = 0.237 x 107 N/C

Using right hand rule , magnetic force is in down direction , hence the electric force has to be opposite to magnetic force that is up direction. so electric field is in down direction since electron experience electric force in opposite direction of electric field.

b)

Using the formula

r = mv/qB

r = (9.1 x 10-31 ) (1.186 x 107) / (1.6 x 10-19x 0.2)

r = 3.37 x 10-4 m

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