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प्रश्न
An electron is accelerated through a potential difference of 5 kV and enters a uniform magnetic field of 0.02 `(wb)/(m^2)` acting normal to the direction of electron motion.
Determine the radius of path.
संख्यात्मक
उत्तर
`v = sqrt(2qV)/m = sqrt(2 xx 1.6 xx 10^(−19) xx 5000)/(9.1 xx 10^(−31)) = 4193 xx 10^4`
`r = (mv)/(qB) =( 9.1 xx 10^(−31) xx 4193 xx 10^4)/( 1.6 xx 10^(−19) xx 0.02) = 0.0119 m`
Radius of path = 0.0119 m
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Motion of Electron in Magnetic Field
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