Question

30. An electron of 40 eV energy is revolving in a circular o
2
path in a magnetic field of 9 x 10-5 T. Determine:
(i) speed of the electron, (ii) radius of the circular path.
Ans. (i) 3.75 x 10 ms (ii) 23.7 cm.
-1​

Answer 1

Answer:

=35.72×10  

−8

s

Explanation:

Magnetic field  B=1×10  

−4

 Wb/m  

2

Mass of electron  m=9.1×10  

−31

 kg

Magnitude of charge of electron  e=1.6×10  

−19

 C

Kinetic energy of electron  K=10eV

(i) : Speed of electron   v=  

m

2K

=  

9.1×10  

−31

2×10×1.6×10  

−19

=5.93×10  

6

 m/s

(ii) :  Radius of circular path    R=  

eB

mv

⟹ R=  

1.6×10  

−19

×10  

−4

9.1×10  

−31

×5.93×10  

6

=33.73×10  

−2

 m=33.73 cm

(iii) :  Time period   T=  

eB

2πm

⟹ T=  

1.6×10  

−19

×10  

−4

2π×9.1×10  

−31

​=35.72×10  

−8

 s