91影视

Acceleration of a moving particle is,

$a=\frac{F}{m}$ ...(i)

There is force experienced by a moving charge in the direction perpendicular to its own velocity and also to the magnetic field. Mathematically this force is given as,

$\mathrm{F}=\left|\mathrm{q}\right|\mathrm{惫叠蝉颈苍蠁}$ ...(ii)

Here, q is the charge, v is the velocity of the moving charge, B is the strength of the magnetic field and 萤 is the angle between the motion of the charge particle and the magnetic field.

The magnetic field in the vicinity of a current carrying wire is,

$\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟿蟿谤}}$

Both the direction of the force and the magnetic field can be inferred from right hand rule.

Substitute and rearrange equation (i),

$\begin{array}{r}\mathrm{a}=\frac{\left|q\right|\mathrm{vBsin}鈦�蠁}{m}\\ =\frac{\mathrm{ev}}{m}\left(\frac{{渭}_{0}l}{2蟺r}\right)\end{array}$

Substitute all the values in the above equation,

$\begin{array}{r}\mathrm{a}=\frac{\left(1.6脳{10}^{鈭�19}\mathrm{C}\right)\left(2.50脳{10}^5\mathrm{m}/\mathrm{s}\right)\left(4\mathrm{蟺}脳{10}^{鈭�7}\mathrm{T}脳\mathrm{m}/\mathrm{A}\right)(13.0\mathrm{A})}{\left(9.11脳{10}^{鈭�31}\mathrm{kg}\right)\left(2\mathrm{蟺}\right)(0.0200\mathrm{m})}\\ =5.7脳{10}^{12}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the magnitude of acceleration is $5.7脳{10}^{12}\mathrm{m}/{\mathrm{s}}^2$and is directed away from the wire.

To maintain the trajectory of the particle, the electric force must balance out the magnetic force.

$\begin{array}{r}\mathrm{eE}=\mathrm{evB}\\ \mathrm{E}=\mathrm{vB}\\ =\mathrm{v}\frac{{\mathrm{渭}}_0\mathrm{I}}{2\mathrm{蟺谤}}\\ =\frac{(250,000\mathrm{m}/\mathrm{s})\left(4\mathrm{蟺}脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}\right)(13.0\mathrm{A})}{2\mathrm{蟺}(0.0200\mathrm{m})}\\ =32.5\mathrm{N}/\mathrm{C}\end{array}$

Thus, the magnitude of the electric field is 32.5 N/C and it is directed away from the wire.

The weight due to gravity on the electron is,

$\begin{array}{r}\mathrm{mg}=\left(9.1脳{10}^{鈭�31}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ 鈮�{10}^{鈭�29}\mathrm{N}\end{array}$

Force due to the electric field is,

$\begin{array}{r}{\mathrm{F}}_{\mathrm{el}}=\mathrm{eE}\\ =\left(1.6脳{10}^{鈭�19}\mathrm{C}\right)(32.5\mathrm{N}/\mathrm{C})\\ 鈮�5{.10}^{鈭�18}\mathrm{N}\end{array}$

So,$F_{el}=5脳{10}^{11}{F}_{grav}$

Thus, the effects of gravity can be neglected as it is negligible compared to the force due to electric force.