A charged particle fo mass `m` and charge `q` enters along `AB` at point `A` in unifrom magnetic field existing in the rectangular region of size `a xx b`. The particle leaves the region exactly at corner point `C`. What is the speed `v` of the particle?

Options

A. `(qB(a^(2) + b^(2)))/(2mb)`
B. `(qB(a^(2) + b^(2)))/(2ma)`
C. `(qB(a^(2) - b^(2)))/(2ma)`
D. `(qB(a^(2) - b^(2)))/(2mb)`

Answer

A

Explanation

Let speed of the particle is `v` is speed wil remain constant)

`r = AO = CO = (mv)/(qB)` also `(a)/(r) = sin theta, (r-b)/(r) = cos theta`
Solve the above equation to get `v = (qB(a^(2) + b^(2)))/(2mb)`