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NCERT Class XII Chapter
Moving Charges and Magnetism
Questions With Solutions

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Question : 11
Total: 28
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106ms–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
(e = 1.6 × 10–19 C, me = 9.1 × 10–31 kg)
Solution:  
The magnetic force f = qvB act normal to the direction of motion, thus provide the necessary centripetal force to follow the circular path
qvB =
mv2
r
r =
mv
qB
 =
9.1×10−31×4.8×106
1.6×10−19×6.5×10−4
 or r = 4.2 × 10–2 = 42 mm
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