A uniform magnetic field of intensity `1T` is applied in a circular region of radius `0.1 m`, directed into the plane of paper. A charged particle of mass `5xx10^(-5)kg` and charge `q=5xx10^(-4)C` enters the field with velocity `1//sqrt(3)m//s` making an angle of `phi` with a radial line of circular region in such a way that it passes through centre of applied field the angle `phi` is

Options

A. `60^(@)`
B. `30^(@)`
C. `45^(@)`
D. `90^(@)`

Explanation

`r_(1)=(mV)/(qB)=(5xx10^(-5)xx(1//sqrt(3)))/(5xx10^(-5)xx1)= (1)/(10sqrt(3))`
By `sin` rule, `(r_(1))/(sin(90^(@)-phi))=(r )/(sin2phi)`
`(1)/(10sqrt(3)cosphi)=(0.1)/(2sinphicosphi)`
`sinphi=(0.1xx10sqrt(3))/(2)=(sqrt(3))/(2)`
`phi=60^(@)`