Question

A long, straight wire carries a 13.0-A current. An electron is fired parallel to this wire with a velocity of 250 km/s in the same direction as the current, 2.00 cm from the wire.

(a) Find the magnitude and direction of the electron’s initial acceleration.

(b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire? (c) Is it necessary to include the effects of gravity? Justify your answer.

Short answer

(a) The magnitude and direction of the initial acceleration of the electron is $5.7\times {10}^{12}\mathrm{m}/{\mathrm{s}}^2$.

(b) The magnitude and direction of uniform electric field that will allow the electron to continue to travel parallel to the wire is 32.5 N/C .

(c) The effects of gravity can be neglected.

Step-by-step solution

(a) Determination of the magnitude and direction of the electron’s initial acceleration.

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,

$\mathrm{F}=\left|\mathrm{q}\right|\mathrm{vBsin\phi }$ ...(ii)

Here, q is the charge, v is the velocity of the moving charge, B is the strength of the magnetic field and ө 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,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\tau \tau r}}$

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

Substitute and rearrange equation (i),

$\begin{array}{r}\mathrm{a}=\frac{\left|q\right|\mathrm{vBsin}\phi }{m}\\ =\frac{\mathrm{ev}}{m}\left(\frac{{\mu }_{0}l}{2\pi r}\right)\end{array}$

Substitute all the values in the above equation,

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

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

(b) Determination of the magnitude and direction of uniform electric field that will allow the electron to continue to travel parallel to the wire.

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

$\begin{array}{r}\mathrm{eE}=\mathrm{evB}\\ \mathrm{E}=\mathrm{vB}\\ =\mathrm{v}\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}\\ =\frac{(250,000\mathrm{m}/\mathrm{s})\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)(13.0\mathrm{A})}{2\mathrm{\pi }(0.0200\mathrm{m})}\\ =32.5\mathrm{N}/\mathrm{C}\end{array}$

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

(c) Determination of the necessity to include the effects of gravity.

The weight due to gravity on the electron is,

$\begin{array}{r}\mathrm{mg}=\left(9.1\times {10}^{-31}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ \approx {10}^{-29}\mathrm{N}\end{array}$

Force due to the electric field is,

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

So,$F_{el}=5\times {10}^{11}{F}_{grav}$

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