Question

Figure P28.62shows an end view of two long, parallel wires perpendicular to the xy-plane, each carrying a current Ibut in opposite directions. (a) Copy the diagram, and draw vectors to show the field of each wire and the net $\overrightarrow{\mathbf{B}}$ field at point P. (b) Derive the expression for the magnitude of at any point on the x-axis in terms of the x-coordinate of the point. What is the direction of $\overrightarrow{\mathbf{B}}$ ? (c) Graph the magnitude of at points on the x-axis. (d) At what value of is the magnitude of a maximum? (e) What is the magnitude of $\overrightarrow{\mathbf{B}}$when x >>a ?

Short answer

(a) The required vectors are drawn.

(b) The expression of the $\overrightarrow{\mathrm{B}}$is $\frac{{\mu }_{0}la}{\pi \left(x^2+a^2\right)}$and its directed towards positive x-direction.

(c) The required graph is shown.

(d) The value of x where the magnitude of $\overrightarrow{\mathrm{B}}$is maximum is zero.

(e) The magnitude of $\overrightarrow{\mathrm{B}}$when is greater than a is $\mathrm{B}=\frac{{\mu }_0\mathrm{la}}{\pi {\mathrm{x}}^2}$.

Step-by-step solution

Step 1:

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,

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

Here, B is the magnetic field, ${\mathbf{\mu }}_{\mathbf{0}}$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.

(b) Determination of the expression and direction of the .

At a given point on the x-axis,

The net magnetic field is,

$\begin{array}{r}{\mathrm{B}}_{\text{net}}=2\mathrm{Bsin}\mathrm{\theta }\\ {\mathrm{B}}_{\text{net}}=2\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}\sin \mathrm{\theta }\end{array}$

….. (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,

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

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

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

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

(c) Depiction of the required graph.

The graph B vs x/a is shown below.

(d) Determination of the value of x where the magnitude of is B→ maximum.

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

The magnetic field, $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi a}}$, which is maximum.

(e) Determination of the magnitude of B→ when x is greater than a.

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

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{la}}{{\mathrm{\pi x}}^2}$