91Ó°ÊÓ

A particle of charge qenters a region of uniform magnetic field pointing into the page.

The field deflects the particle a distance$d$ above the original line of flight.

The force on a charge $q$ moving with a velocity $\stackrel{→}{v}$ in the presence of a magnetic field $\stackrel{→}{B}$ is

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{q}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{)}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}\mathbf{.}\mathbf{....}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The momentum of such a particle moving in a circular trajectory of radius$R$ is

$\mathbf{p}\mathbf{=}\mathbf{qBR}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}\mathbf{.}\mathbf{....}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Since $\stackrel{→}{V}$ is towards the right and $\stackrel{→}{B}$points into the page, from equation (1) the force must be pointing upwards if the charge is positive. That is indeed the direction of the deflection. Hence the charge is positive.

In the figure, using Pythagoras theorem,

role="math" localid="1657687257870"

Substitute this in equation (2) and get

$p=qB\frac{a^2+d^2}{2d}$

Thus, the momentum of the particle is $qB\frac{a^2+d^2}{2d}$.