Question

Consider a planeloop of wire that carries a steady current I;we

want to calculate the magnetic field at a point in the plane. We might as well take

that point to be the origin (it could be inside or outside the loop). The shape of the

wire is given, in polar coordinates, by a specified function $\mathbf{r}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$(Fig. 5.62).

(a) Show that the magnitude of the field is

role="math" localid="1658927560350" $\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_0\mathbf{I}}{4\pi }\mathbf{\oint }\frac{}{}$(5.92)

(b) Test this formula by calculating the field at the center of a circular loop.

(c) The "lituus spiral" is defined by a

$\mathbf{r}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\frac{a}{\sqrt{\theta }}\mathbf{0}\mathbf{<}\mathbf{\theta }\mathbf{\le }\mathbf{2}\mathbf{\pi }$

(for some constant a).Sketch this figure, and complete the loop with a straight

segment along the xaxis. What is the magnetic field at the origin?

(d) For a conic section with focus at the origin,

$\mathbf{r}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\frac{p}{1+\mathrm{ecos\theta }}$

where pisthe semi-latus rectum (the y intercept) and eis the eccentricity (e= 0

for a circle, 0 < e< 1 for an ellipse, e= 1 for a parabola). Show that the field is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_0\mathbf{I}}{2p}$regardless of the eccentricity.

Short answer

(a) The magnitude of the magnetic field of a closed plane loop carrying current $I$ is $B=\frac{{\mu }_{0}I}{4\pi }\oint \frac{d\theta }{r}$.

(b) If the loop is a circle, the field at its center is$\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}$.

(c) If the loop has formrole="math" localid="1658927823158" $r\left(\theta \right)=\frac{a}{\sqrt{\theta }}$ , the magnitude of the field is$\frac{{\mu }_{0}I\sqrt{2\pi }}{3a}$. The sketch for the loop is obtained.

(d) If the loop has form $r\left(\theta \right)=\frac{p}{1+e\cos \theta }$, the magnitude of the field is $\frac{{\mu }_{0}I}{2p}$.

Step-by-step solution

Given data

(a) There is a closed loop carrying current$I$.

(c) The loop has trajectory $\mathrm{r}\left(\mathrm{\theta }\right)=\frac{\mathrm{a}}{\sqrt{\mathrm{\theta }}}0<\mathrm{\theta }\le 2\mathrm{\pi }$.

(d) The loop has trajectory$r\left(\mathrm{\theta }\right)=\frac{\mathrm{p}}{1+\mathrm{ecos\theta }}$.

Determine magnetic field of a current carrying wire

The magnetic field of a current carrying wire$I$is:

$\overrightarrow{\mathbf{B}}\mathbf{=}\frac{{\mathbf{\mu }}_0\mathbf{I}}{4\pi }\int \frac{d\overrightarrow{l}\times \widehat{r}}{{\mathbf{r}}^2}$ …… (1)

Here, ${\mathbf{\mu }}_0$ is the permeability of free space and $\mathit{d}\overrightarrow{\mathbf{l}}$ is an infinitesimal length element on the wire.

Determine magnetic field of a closed plane current carrying loop

(a)

For the given configuration,

$|\mathrm{d}\overrightarrow{\mathrm{l}}\times \widehat{\mathrm{r}}|=\mathrm{rd\theta }$

Thus, from equation (1) write as:

$B=\frac{{\mu }_{0}I}{4\pi }\oint \frac{d\theta }{r}$ …… (2)

Determine the magnetic field at the center of a current carrying circular loop

(b)

For a circular loop,$r=R$. Here,$R$is the radius of the loop.

Thus, from equation (2),

$\begin{array}{c}B=\frac{{\mu }_{0}I}{4\pi R}\oint d\theta \\ =\frac{{\mu }_{0}I}{2R}\end{array}$

This agrees with the field at the center of a current carrying loop.

Determine the magnetic field for a loop defined by   r(θ)=aθ

(c)

The sketch of $\mathrm{r}\left(\mathrm{\theta }\right)=\frac{\mathrm{a}}{\sqrt{\mathrm{\theta }}}0<\mathrm{\theta }\le 2\mathrm{\pi }$ is given in the following figure.

Thus, from equation (2) is as follows:

Thus, the field is $\frac{{\mathrm{\mu }}_0\mathrm{I}\sqrt{2\mathrm{\pi }}}{3\mathrm{a}}$.

Determine the magnetic field of a conic section

(d)

The field for$\mathrm{r}\left(\mathrm{\theta }\right)=\frac{\mathrm{p}}{1+\mathrm{ecos\theta }}$is obtained from equation (2) as:

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\oint \frac{1+\mathrm{ecos\theta }}{\mathrm{p}}\mathrm{d\theta }\\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi p}}\underset{0}{\overset{2\mathrm{\pi }}{\int }}(1+\mathrm{ecos\theta })\mathrm{d\theta }\\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi p}}{[\mathrm{\theta }-\mathrm{esin\theta }]}_0^{2\mathrm{\pi }}\\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{p}}\end{array}$

Thus, the field is obtained as: $\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{p}}$.