A long, straight wire carries a current of 8.60 A. An electron is travelling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and travelling at a speed of $\mathbf{6}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{m}\mathbf{/}\mathbf{s}$ directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron?

A vector field that describes the influence of magnetic nature on electric charges in motion, conductors carrying current and other materials that are characterized as magnetic is known as the magnetic field.

There is the 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,

$F=\left|q\right|vBsin\theta$ ...(i)

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 charged particle and the magnetic field.

Given data:

• $r=0.0450\mathrm{m}$
• $\theta ={90}^{\circ }$
• $I=8.60\mathrm{A}$
• $v=6.00\times {10}^4\mathrm{m}/\mathrm{s}$
• ${\mu }_0=4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}$
• $e=1.60\times {10}^{-19}\mathrm{C}$

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

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

Substitute q = e and B in equation (i),

$F=\frac{ev{\mu }_{0}I\sin \theta }{2\pi r}$

Substitute the values as:

$\begin{array}{r}F=\frac{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(6.00\times {10}^4\mathrm{m}/\mathrm{s}\right)\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(8.60\mathrm{A}\right)\left(\sin {90}^{\circ }\right)}{2\pi \left(0.0450\mathrm{m}\right)}\\ =3.67\times {10}^{-19}\mathrm{N}\end{array}$

The direction of the force can be inferred from the right-hand rule and is in the same direction as the current, and the magnitude is $3.67\times {10}^{-19}\mathrm{N}$