A particle accelerated by a potential difference `V` flies through a uniform transverse magnetic field with induction B. The field occupies a region of space d in thickness. Prove that the angle a through which the particle deviates from the initial direction of its motion is given by.
`alpha=sin^(-1)(dBsqrt((q)/(2 Vm)))`
where `m` is the mass of the particle.

Options

A. `theta=sin^(-1)(dBsqrt((qV)/(2m)))`
B. `theta=sin^(-1)(dBsqrt((q)/(2mV)))`
C. `theta=tan^(-1)(dBsqrt((qV)/(2m)))`
D. `theta=tan^(-1)(dBsqrt((q)/(2mV)))`

Answer

B