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प्रश्न
An electron is moving with a speed of 3.2 × 107 m/s in a magnetic field of 6.00 × 10-4 T perpendicular to its path. What will be the radium of the path? What will be frequency and the energy in keV?
[Given: mass of electron = 9.1 × 10−31 kg, charge e = 1.6 × 10−19 C, 1 eV = 1.6 × 10−19 J]
उत्तर
Given: v = 3 × 10-7 m/s, B = 6 x 10-4 T,
me = 9 x 10-31 kg, e = 1.6 x 10-19 C,
1 eV = 1.6 x 10-19 J
The radius of the circular path,
r = `("m"_"e""v")/(|"e"|"B")`
`= ((9 xx 10^-31)(3 xx 10^7))/((1.6 xx 10^-19)(6 xx 10^-4)) = 2.7/9.6` = 0.2812 m
The frequency of revolution,
f = `(|"e"|"B")/(2pi"m"_"e")`
`= ((1.6 xx 10^-19)(6 xx 10^-4))/(2 xx 3.142 xx (9 xx 10^-31))`
`= 9.6/(18 xx 3.142) xx 10^8 = 16.97`MHz
Since the magnetic force does not change the kinetic energy of the charge,
KE = `1/2 "m"_"e" "v"^2 = 1/2 (9 xx 10^-31)(3 xx 10^7)^2 = 81/2 xx 10^-17`J
`= 81/(2(1.6 xx 10^-19)) xx 10^-17` eV
`= 8.1/3.2 xx 10^3`
= 2.531 ke V
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