91影视

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,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{蟺谤}}$

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

(a) Depiction of the vectors drawn on the given figure.

At a given point on the x-axis,

The net magnetic field is,

$\begin{array}{r}{\mathrm{B}}_{\text{net}}=2\mathrm{Bsin}鈦�\mathrm{胃}\\ {\mathrm{B}}_{\text{net}}=2\frac{{\mathrm{渭}}_0\mathrm{I}}{2\mathrm{蟺谤}}\sin 鈦�\mathrm{胃}\end{array}$

鈥�.. (1)

Now, refer to the image in part (a), r is the distance (hypotenuse) or the line joining the point 1 and the point where the Bnetmanifests.

So,$\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}$

Also,

$\begin{array}{r}\sin 鈦�胃=\frac{\text{height}}{\text{hypotenuse}}\\ =\frac{a}{r}\\ =\frac{a}{\sqrt{x^2+a^2}}\end{array}$

Substitute above both equations for r and sin$胃$ into equation (1), and you have

$\begin{array}{r}\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{I}}{\mathrm{蟺}\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\frac{\mathrm{a}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\\ =\frac{{\mathrm{渭}}_0\mathrm{la}}{\mathrm{蟺}\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}\end{array}$

Hence, this magnetic field acts in the positive x-direction.

The graph B vs x/a is shown below.

When x=0, i.e. at the origin,

The magnetic field, $\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{I}}{\mathrm{蟺补}}$, which is maximum.

From part (b), the magnitude of B at x>>a is,

$\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{la}}{{\mathrm{蟺虫}}^2}$