1.

A positively ahargedparticel (q, m) enters in a uniform magnetic field 'B' at an angle of 60^(@) as shown in figure with speed v_(o). Collision between the charge particle and the wall is perfectly elastic. Then find the time in which the charge particel comes out from the magnetic field. ("take"d = ((sqrt3 - 1)mv)/(2qB))

Answer»

`(pim)/(2qB)`
`(pim)/(6qB)`
`(pim)/(3qB)`
`(2pi m)/(3qB)`

Solution :`R cos theta - R//2 = d = (R (SQRT3 - 1))/(2)`
`cos theta - (1)/(2) = (sqrt3)/(2) - (1)/(2)`
`theta = 30^(@)`
So, total angle rotates inside the magnetic FIELD `= 30^(@) + 30^(@) = 60^(@)`
Time period `= (m)/(qb) (pi)/(3) = (pi m)/(3qB)`


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