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प्रश्न
A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 106 ms-1. It is projected perpendicularly into the magnetic field of strength 0.2 T. The radius of the circle described is ______ cm.
विकल्प
11.00
12.00
13.00
14.00
MCQ
रिक्त स्थान भरें
उत्तर
A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 106 ms-1. It is projected perpendicularly into the magnetic field of strength 0.2 T. The radius of the circle described is 12.00 cm.
Explanation:
`R = (mv)/(qB)`
`q xx 12 xx 10^3 = 1/2m xx (10^6)^2`
`(24 xx 10^3)/10^12 = m/q`
R = `(24 xx 10^3 xx 10^6)/(10^12 xx 0.2)`
R = `12 xx 10^-2` m
R = 12 cm
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Motion in a Magnetic Field
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