AB and AC are boundary lines within which a magnetic field B exists. If the magnetic field is absent, a charged particle of mass m and charge q must have passed through a point P on angle bisector of -BAC, at a distance `r(sqrt2)` from A, if it has fallen on AB normally at a point Q such that QP =r. If the magnetic field is present, how much time the charged particle will take to come out of the magnetic field

Options

A. `(m pi)/(Bq)`
B. `(2m pi)/(Bq)`
C. `(4m pi)/(Bq)`
D. `(m pi)/(2Bq)`

Answer

D

Explanation

Since, `QP =r, AP=rsqrt(2) and /_AQP=90^(@)`
`AQ = r and /_QAP = 45^(@)`
The same logic concludes `AR=r` and `/_PAR=45^(@)`
so, `/_QAR = 90^(@)`
`r=(mv)/(Bq)`
Distance travelled `=(pi r)/(2)` ltbRgt Time` =(pi)/(2)* (mv)/(Bq)*1/v = (m pi)/(2Bq)`.