91Ó°ÊÓ

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

(c) The loop has trajectory $\mathrm{r}\left(\mathrm{θ}\right)=\frac{\mathrm{a}}{\sqrt{\mathrm{θ}}}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0<\mathrm{θ}≤2\mathrm{Ï€}$.

(d) The loop has trajectory$r\left(\mathrm{θ}\right)=\frac{\mathrm{p}}{1+\mathrm{\pm 𳦴Dzõθ}}$.

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

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\frac{{\mathbf{μ}}_0\mathbf{I}}{4\mathrm{π}}∫\frac{d\stackrel{→}{l}×\widehat{r}}{{\mathbf{r}}^2}$ …… (1)

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

(a)

For the given configuration,

$|\mathrm{d}\stackrel{→}{\mathrm{l}}×\widehat{\mathrm{r}}|=\mathrm{°ù»åθ}$

Thus, from equation (1) write as:

$B=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{Ï€}}âˆ\circledR \frac{d\mathrm{θ}}{r}$ …… (2)

(b)

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

Thus, from equation (2),

$\begin{array}{c}B=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{Ï€}R}âˆ\circledR d\mathrm{θ}\\ =\frac{{\mathrm{μ}}_{0}I}{2R}\end{array}$

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

(c)

The sketch of $\mathrm{r}\left(\mathrm{θ}\right)=\frac{\mathrm{a}}{\sqrt{\mathrm{θ}}}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0<\mathrm{θ}≤2\mathrm{Ï€}$ is given in the following figure.

Thus, from equation (2) is as follows:

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

(d)

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

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€}}âˆ\circledR \frac{1+\mathrm{\pm 𳦴Dzõθ}}{\mathrm{p}}\mathrm{»åθ}\\ =\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€\pm è}}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}(1+\mathrm{\pm 𳦴Dzõθ})\mathrm{»åθ}\\ =\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€\pm è}}{[\mathrm{θ}−\mathrm{\pm ð²õ¾\pm ²Ôθ}]}_0^{2\mathrm{Ï€}}\\ =\frac{{\mathrm{μ}}_0\mathrm{I}}{2\mathrm{p}}\end{array}$

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