Flower-like loop

According to Biot-Savart's law: $d\overrightarrow{B_0}=\frac{\mu_0}{4\pi} \frac{I_0 \overrightarrow{dl} \times \overrightarrow{r}}{r^3}$ .

Therefore, $dB_0=\frac{\mu_0}{4\pi} \frac{I_0 (r d\theta) r}{r^3}=\frac{\mu_0}{4\pi} \frac{I_0 d\theta}{r}=\frac{\mu_0 I_0}{4\pi R} \frac{d\theta}{1+\varepsilon \cos (n\theta)}$

Therefore, $B_0=\int\limits_0^{2\pi} \frac{\mu_0 I_0}{4\pi R} \frac{d\theta}{1+\varepsilon \cos (n\theta)}$ .

$B_0=\frac{\mu_0 I_0}{2R \sqrt{1-\varepsilon^2}}=\frac{B}{\sqrt{1-\varepsilon^2}}=1.15(mT)$ .

The Biot-Savart equation is given by:

$\vec{B} = \frac{\mu_0 I_0}{4 \pi} \int_{C} \frac{\vec{dl} \times \hat{r}}{r^2}$

We can write the line element of a circuit in the $z=0$ plane in cylindrical coordinates as:

$\vec{dl} = dr \;\hat{r} + r\; d\theta \; \hat{\theta}$

Then the cross product $\vec{dl} \times \hat{r}$ in the Biot-Savart law is:

$\vec{dl} \times \hat{r} = \begin{vmatrix} \hat{r} & \hat{\theta} & \hat{z} \\ dr &r\; d\theta &0 \\ 1 & 0 & 0 \end{vmatrix} = - r\; d\theta \; \hat{z}$

$\vec{B} = \frac{\mu_0 I_0}{4 \pi} \int_{C} \frac{- r\; d\theta \; \hat{z}}{r^2}$

$\left|\vec{B}\right| = \frac{\mu_0 I_0}{4 \pi} \int_{C} \frac{d\theta}{r}$

In the circular loop case $r(\theta) = R$ :

$\left|\vec{B_1}\right| = \frac{\mu_0 I_0}{4 \pi} \int_{0}^{2\pi} \frac{d\theta}{R}$

$\left|\vec{B_1}\right| = \frac{\mu_0 I_0}{2 R}$

In the flower-like curve case $r(\theta) = R(1+\varepsilon cos(n\theta))$

$\left|\vec{B_2}\right| = \frac{\mu_0 I_0}{4 \pi} \int_{0}^{2\pi} \frac{d\theta}{R(1+\varepsilon cos(n\theta))}$

$\left|\vec{B_2}\right| = \frac{\mu_0 I_0}{4 R \pi} \int_{0}^{2\pi} \frac{d\theta}{1+ \frac{cos(5\theta)}{2}}$

$\left|\vec{B_2}\right| = \frac{\mu_0 I_0}{4 R \pi} \frac{4 \pi}{\sqrt{3}}$

$\left|\vec{B_2}\right| = \frac{\mu_0 I_0}{2 \pi R} \frac{2\sqrt{3}}{3}$

$\left|\vec{B_2}\right| = \left|\vec{B_1}\right| \frac{2\sqrt{3}}{3}$

$\left|\vec{B_2}\right| = 1\; mT \; \times \frac{2\sqrt{3}}{3}$

$\left|\vec{B_2}\right| = 1.15 \; mT$

Let us consider a small current carrying element at an angle $\theta$ at a distance $r$ from center.Let us assume center to be $(0,0)$ and let the coordinates of this element be $(x_{0},y_{0})$ . Now let us extend this element as a big line and let $p$ denote the length of perpendicular dropped on this line from origin.Since, the small magnetic fields of these elements are in same direction, we can simply add their magnitudes.

$|d\vec{B}| = \frac{\mu_{0}I_{0}|d\vec{l} \times \vec{r}|}{4 \pi r^3} = \frac{\mu_{0}I_{0}pdl}{4 \pi r^3}$ ... $(i)$

Let $f^{'}(x)$ denote the slope of this element.

By simple trigonometry , $dl = dx\sqrt{1 + (f^{'}(x))^2}$ ..... $(ii)$

Now, let us write the equation of extended line :

$y - y_{0} = (x - x_{0})f^{'}(x)$ ,

Hence , $p = \frac{|y_{0} - x_{0}f^{'}(x)|}{\sqrt{1 + (f^{'}(x))^2}}$ ... $(iii)$

From $(ii)$ and $(iii)$ , we get :

$pdl = |y_{0}dx - x_{0}dy|$ ,

Here, use $x_{0} = rcos\theta$ , and $y_{0} = rsin\theta$ , take derivative and simplify to get

$pdl = r^2d\theta$ . Put this value in $(i)$ to get :

$|d\vec{B}| = \frac{\mu_{0}I_{0}d\theta}{4 \pi r} = \frac{\mu_{0}I_{0}d\theta}{4 \pi R(1 + \eta cos(n\theta))}$

$\Rightarrow B = \frac{B_{circle}}{2 \pi} \int_{0}^{2\pi} \frac{d\theta}{1 + \eta cos(n\theta)}$ = $1.15B = \boxed{1.15 mT}$

For a current $I_0$ flowing in a curve $C$ , the magnetic field at the point $\mathbf{r}$ is $\mathbf{B}(\mathbf{r}) \; = \; \frac{\mu_0 I_0}{4\pi}\oint_C \frac{d\mathbf{s} \times (\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3}$ Thus the field at the origin is $\mathbf{B}(\mathbf{0}) \; = \; \frac{\mu_0I_0}{4\pi}\oint_C \frac{\mathbf{s}\times d\mathbf{s}}{|\mathbf{s}|^3} \; = \; \frac{\mu_0I_0}{4\pi}\int_0^{2\pi} \frac{d\theta}{s} \mathbf{k}$ where $\mathbf{k}$ is a unit vector normal to the plane.

For the circular loop the magnitude of the magnetic field at $O$ is $B_1 \; = \; \frac{\mu_0I_0}{4\pi} \int_0^{2\pi} \frac{d\theta}{R} \; = \; \frac{\mu_0I_0}{2R} \; = \; 1 \mbox{ mT}$ For the flower-like loop the magnitude of the magnetic field at $O$ is $\begin{array}{rcl} B_2 & = & \frac{\mu_0I_0}{4\pi} \int_0^{2\pi} \frac{d\theta}{R(1 + \varepsilon\cos n\theta)} \\ & = & \frac{\mu_0I_0}{4n\pi R} \int_0^{2n\pi}\frac{d\theta}{1 + \varepsilon\cos\theta} \\ & = & \frac{\mu_0I_0}{4\pi R}\int_0^{2\pi}\frac{d\theta}{1 + \varepsilon\cos\theta} \; = \; \frac{\mu_0I_0}{2R\sqrt{1-\varepsilon^2}} \end{array}$ or $\frac{1}{\sqrt{1-\varepsilon^2}} = \frac{2}{\sqrt{3}} = 1.15$ mT.