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NCERT Class XII Chapter
Moving Charges and Magnetism
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Question : 11 of 28
Marks: +1, -0
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10−410^{-4} T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s−110^{6}\,\mathrm{m}\,\mathrm{s}^{-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 × 10−1910^{-19} C, mem_e = 9.1 × 10−3110^{-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 = mv2r\frac{mv^2}{r}
r = mvqB\frac{mv}{qB} = 9.1×10−31×4.8×1061.6×10−19×6.5×10−4\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}} or r = 4.2 × 10−210^{-2} = 42 mm
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