Question: A long straight wire carries a current of $\mathbf{50}\mathbf{ }\mathbf{\text{A}}$. An electron, traveling at$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$, is$\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{ }\mathbf{\text{m}}$from the wire. What is the magnitude of the magnetic force on the electron if the electron velocity is directed (a) toward the wire, (b) parallel to the wire in the direction of the current, and (c) perpendicular to the two directions defined by (a) and (b)?

1. The current is,$i=50 \text{A}$
2. The speed of electron is,$V=1\times {10}^7 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$
3. The distance is,$r=0.05 \text{m}$

We use the formula of magnetic field due to an infinitely long wire at perpendicular distance to calculate the magnetic field due to the wire andtaking the cross product of velocity and magnetic field, we can find the direction of magnetic force and its magnitude.

Formula:

${\overrightarrow{\mathbf{F}}}_{\mathbf{B}}\mathbf{=}\mathit{q}\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$

The magnetic force on the charged particles is given as

${\overrightarrow{\mathbf{F}}}_{\mathbf{B}}\mathbf{=}\mathit{q}\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

The charged particle is an electron; hence, the force on the electron due to magnetic field is given by

${\overrightarrow{\mathbf{F}}}_{\mathbf{e}}\mathbf{=}\mathbf{-}\mathit{e}\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

When velocity is towards the wire:

The velocity of electron is in the downward direction.We have a current flow in the +x direction, and electron at$r=0.05\text{m}$above the wire (where “up” is the +y direction). Thus, the field produced by the current points in the +z direction at P.

$\overrightarrow{V}=-V\hat{\mathrm{j}}$

$V=1.0\times {10}^7 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

The magnitude of magnetic field due to a long straight wire at perpendicular distance$r$is given by

$B=\frac{{\mu }_{0}i}{2\pi r}$

The magnetic field is along $+Z$direction, hence, in vector form

$\overrightarrow{B}=\frac{{\mu }_{0}i}{2\pi r}\hat{k}$

Substituting the value of the magnetic field and velocity in equation-

${\overrightarrow{F}}_e=-e\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

${\overrightarrow{F}}_e=-e\left(V\hat{j}\times \frac{{\mu }_{0}i}{2\pi r}\hat{k}\right)$ …(2)

${\overrightarrow{F}}_e=-eV\frac{{\mu }_{0}i}{2\pi r}\left(\hat{j}\times \hat{k}\right)$

${\overrightarrow{F}}_e=-eV\frac{{\mu }_{0}i}{2\pi r}\hat{\mathrm{i}}$

Now substituting all values in the above, we get

${\overrightarrow{F}}_e=-\frac{\left(-1.6\times {10}^{-19} \text{C}\right)\left(1.26\times {10}^{-6} {\displaystyle \raisebox{1ex}{$\text{T.m}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.}\right)\left(1\times {10}^7 {\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.}\right)\left(50 \text{A}\right)}{2\pi \times 0.05 \text{m}}\hat{\mathrm{i}}$

${\overrightarrow{F}}_e=\left(3.2\times {10}^{-16}\mathrm{N}\right)\hat{\mathrm{i}}$

And the magnitude is given by role="math" localid="1662828889670" $F_e=3.2\times {10}^{-16} \text{N}$.

The velocity is along +x direction. But the magnetic field is in the same direction.

$$\begin{gathered} \overrightarrow{V}=V\hat{\mathrm{i}} \\ =\left(3.2\times {10}^{-16} \text{N}\right)\hat{\mathrm{i}} \end{gathered}$$

Hence, from relation 2, we get

role="math" localid="1662829605536" $$\begin{gathered} {\overrightarrow{F}}_e=-e\left(V\hat{i}\times \frac{{\mu }_{0}i}{2\pi r}\hat{k}\right) \\ =-eV\frac{{\mu }_{0}i}{2\pi r}\left(\hat{i}\times \hat{k}\right) \\ =-eV\frac{{\mu }_{0}i}{2\pi r}\left(-\hat{j}\right) \\ =eV\frac{{\mu }_{0}i}{2\pi r}\hat{j} \end{gathered}$$

Substitute all the value in the above equation.

role="math" localid="1662830189646" $$\begin{gathered} \overrightarrow{F_e}=\frac{\left(1.6\times {10}^{-19} \text{C}\right)\left(1.26\times {10}^{-6}{\displaystyle \raisebox{1ex}{$ \text{T.m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}\right)\left(1\times {10}^7 {\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.}\right)\left(50 \text{A}\right)}{2\pi \times 0.05 \text{m}}\hat{\mathrm{j}} \\ {\overrightarrow{F}}_e=\left(3.2\times {10}^{-16} \text{N}\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of the force is the same as result (a) but direction of force is along +y direction.

In this case, the velocity is in both $\pm Z$direction.

From relation 2

${\overrightarrow{F}}_e=-e\left(\pm V\hat{k}\times \frac{{\mu }_{0}i}{2\pi r}\hat{k}\right)$

It will be zero because$\left(\hat{k}\times \hat{k}\right)=0$

Hence, there will be no force on the electron.