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प्रश्न
A particle of charge 2.0 × 10−8 C and mass 2.0 × 10−10 g is projected with a speed of 2.0 × 103 m s−1 in a region with a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.
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उत्तर
Given:
Charge of the particle, q = 2.0 × 10−8 C
Mass of the particle, m = 2.0 × 10−10 g
Projected speed of the particle, v = 2.0 × 103 m s−1
Uniform magnetic field, B = 0.10 T.
As per the question, the velocity is perpendicular to the field.
So, for the particle to move in a circle,the centrifugal force to the particle will be provided by the magnetic force acting on it.
Using qvB =`(mv^2)/(r)` , where r is the radius of the circle formed,
`r = (mv)/(qB)`
= `(2xx10^-13xx2xx10^3)/(2xx10^-8xx0.10)`
= 20 cm
Time period,
`T = (2pim)/(qB)`
=` (2xx3.14xx2xx10^-13}/(2xx10^-8xx0.10)`
= 6.28 × 10-4 s
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