A charged particle of mass m = 1 mg and charge `q = 1 (mu) C` enter along AB at point A in a uniform magnetic field B = 1.2 T existing in the rectangular region of size `a xx b`, where a = 4 m and b = 3 m. The particle leaves the region exactly at corner point C. What is the speed `v (in m s^(-1))` of the particle?

Answer

5

Explanation

Let speed of the particle is v(speed will remain constant)
`r=AO = CO = (mv).(qB) also a/r = sin theta, (r-b)/(r) = cos theta`
Solve the above equaton to get

`v=qB(a^(2)+b^(2))/(2mb) = 10^(-6) xx 1.2 (4^(2)+3^(2))/(2 xx 10^(-6) xx 3) = 5 m//s`.