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प्रश्न
An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.
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उत्तर
Given:
Kinetic energy of an electron = 100 eV
Radius of the circle = 10 cm
`1/2mv^2` = 100 eV = 1.6 × 10−17 J (1 eV = 1.6 × 10−19 J)
Here,
m is the mass of an electron and v is the speed of an electron. Thus,
1/2 × 9.1 × 10−31 × v2 = 1.6 × 10−17 J
⇒ v2 = 0.35 × 1014
v = 0.591 × 107 m/s
Now,
`r = (mv)/(eB)`
`⇒ B = (mv)/(er)`
`= (9.1xx10^-31xx0.591xx10^7)/(1.6xx10^-19xx0.1)`
B = 3.3613 × 10−4 T
Therefore, the applied magnetic field = 3.4 × 10−4 T
Number of revolutions per second of the electron,
`f = 1/T`
`T = (2pir)/v= 2pim/(eB)`
`T = (2pim)/Be`
`f = (Be)/(2pim)`
`=( 3.4xx10^-4xx1.6xx10^-19)/(2xx3.14xx9.1xx10^-31`
= 0.094 × 108
= 9.4 × 106
f = 9.4 × 106
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