Question

A particle of charge qenters a region of uniform magnetic field $\overrightarrow{\mathbf{B}}$ (pointing intothe page). The field deflects the particle a distanced above the original line of flight, as shown in Fig. 5.8. Is the charge positive or negative? In terms of a, d, Band q,find the momentum of the particle.

Short answer

The charge qentering a region of uniform magnetic field $\overrightarrow{\mathrm{B}}$and getting deflected by a distance $\mathrm{d}$is positive. The momentum of the charge is $\mathrm{qB}\frac{{\mathrm{a}}^2+{\mathrm{d}}^2}{2\mathrm{d}}$.

Step-by-step solution

Given data

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.

Define the formula for the force on a charge in a magnetic field and its momentum

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

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{q}\mathbf{(}\overrightarrow{\mathbf{v}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{)}\mathbf{.}\mathbf{....}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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

$\mathbf{p}\mathbf{=}\mathbf{qBR}\mathbf{.}\mathbf{....}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Determine the momentum of the given charge

Since $\overrightarrow{V}$ is towards the right and $\overrightarrow{B}$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,

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Substitute this in equation (2) and get

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

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