91Ó°ÊÓ

The magnetic field created by the electric field equals the size of that electric current with the same proportions as the free access space.

The magnitude of the magnetic field due to current-carrying wire is,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{Ï€°ù}}$

Here, B is the magnetic field, ${\mathbf{μ}}_{\mathbf{0}}$is the permeability of free space, I is the current, and r is the radius.

Consider the given data as below.

Wire AB carries a current, ${\mathrm{I}}_{\mathrm{wire}}=14.0\mathrm{A}$

The current, I = 5.00 A

Length of the wire, L = 20.0 cm = 0.200 m

The radius, ${\mathrm{r}}_1=10.0\mathrm{cm}=0.100\mathrm{m}$

The radius, ${\mathrm{r}}_{\mathrm{b}}=2.6\mathrm{cm}=0.026\mathrm{m}$

Permeability of free space, ${\mathrm{μ}}_0=4×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}$

From the figure given, the forces on the left and right arm cancel each other. The forces acting on the lower and upper arms are the ones that give a resultant force.

$\begin{array}{r}\mathrm{F}={\mathrm{F}}_{\mathrm{t}}−{\mathrm{F}}_{\mathrm{b}}\\ =\left(\frac{{\mathrm{μ}}_0â‹\dots {\mathrm{I}}_{\text{wire}}}{2\mathrm{Ï€}}\right)\left(\frac{\mathrm{IL}}{{\mathrm{r}}_{\mathrm{t}}}−\frac{\mathrm{IL}}{{\mathrm{r}}_{\mathrm{b}}}\right)\\ =\frac{{\mathrm{μ}}_0â‹\dots \mathrm{IL}â‹\dots {\mathrm{I}}_{\text{wire}}}{2\mathrm{Ï€}}\left(\frac{1}{{\mathrm{r}}_{\mathrm{t}}}−\frac{1}{{\mathrm{r}}_{\mathrm{b}}}\right)\\ \mathrm{Substitute}\mathrm{all}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\\ \mathrm{F}=\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{H}/\mathrm{m}\right)(5.00\mathrm{A})(0.200\mathrm{m})(14.0\mathrm{A})}{2\mathrm{Ï€}}\left(−\frac{1}{0.100\mathrm{m}}+\frac{1}{0.026\mathrm{m}}\right)\\ =7.97×{10}^{−5}\mathrm{N}\end{array}$

Hence, the top end force is directed towards the wire and so the resultant force $7.97×{10}^{−5}\mathrm{N}$ is concluded to be towards the wire.