Question

An electron beam passes through a magnetic field of 2 x103 T and an electric field of 3.4 x 10¹ Vm¹ both acting simultaneously. If the path of the electron remains undeviated, calculate the speed of the electrons. If the electric field is removed, what will be the radius of the circular path? Mass of electron=9.1×10-3¹ Kg​

Answer 1

Answer:

B=2×10

−3

Wb/m

2

,

E=1×10

4

V/m

2

Since the path of electron remains undeviated, qvB=qE or

v=

B

E

=

2×10

−3

1×10

+4

=0.5×10

7

=5×10

6

m/s

If the electric field is removed, the path of the charged particle is circular and magnetic field provides the necessary centripetal force, i.e.,

r

mv

2

=Bev⇒r=

Be

mv

=

2×10

−3

×1.6×10

−19

9.1×10

−31

×5×10

6

=14.3×10

−3

m=1.43cm

Explanation:

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