91Ó°ÊÓ

The radius,$R_b=\raisebox{1ex}{${\displaystyle R_a}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

You can calculate the magnetic fields at the center of both the circular coils and then you can take their ratio. The magnitude of the magnetic dipole moment can be defined as the product of number of turns, current in that loop and the area of that loop. You can calculate the magnetic dipole moments of both the coils and take their ratio.

Formulae:

The magnetic dipole moment is,

$\mathrm{μ}=NiA$

Here,$N$is the number of turns,$i$is the current,$A$is the area.

The magnetic field is,

localid="1663240548326" $B=\frac{{\mathrm{μ}}_{0}i{R}^2}{2{(R^2+Z^2)}^{\frac{3}{2}}}$

Here,${\mathrm{μ}}_0$is the permeability of free space having a value $4\mathrm{π}×{10}^{-7}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^{\text{2}}$}\right.$,$R$is the radius,$Z$and is the distance.

The magnetic field due to a circular wire at the point $P$at distance $Z$is given by equation.

localid="1663240563016" $B=\frac{{\mathrm{μ}}_{0}i{R}^2}{2{\left(R^2+Z^2\right)}^{\frac{3}{2}}}$

For magnetic field at the center, $Z=0$, so the equation becomes,

$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_{0}i{R}^2}{2R^3}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2R}\end{array}$

The magnetic field for $a$is

$B_a=\frac{{\mathrm{μ}}_{0}i}{2R_a}$

As the coil $b$has two loops, so the magnetic field for $b$is,

$B_b=\frac{2{\mathrm{μ}}_{0}i}{2R_b}$

By taking ratio $\raisebox{1ex}{$B_b$}\!\left/ \!\raisebox{-1ex}{$B_a$}\right.$you get,

$$\begin{gathered} \frac{B_b}{B_a}=\frac{2{\mathrm{μ}}_{0}i}{2R_b}×\frac{2R_a}{{\mathrm{μ}}_{0}i} \\ \frac{B_b}{B_a}=\frac{2R_a}{R_b} \end{gathered}$$

But from initial condition,$R_b=\raisebox{1ex}{${\displaystyle R_a}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\begin{array}{rcl}\frac{B_b}{B_a}& =& \frac{2R_a}{{\displaystyle \frac{R_a}{2}}}\\ & =& 4\end{array}$

Hence, the ratio is$4$.

The magnetic dipole moment is given as,

$\mathrm{μ}=NiA$

The magnetic dipole moments for $a$and $b$are as follow.

${\mathrm{μ}}_a=i{A}_a$

${\mathrm{μ}}_b=2i{A}_b$

By taking the ratio of the dipole moment magnitudes of the coils, you have

$\frac{{\mathrm{μ}}_b}{{\mathrm{μ}}_a}=\frac{2i{A}_b}{i{A}_a}$

Since the area of circular loop is,

$A=\mathrm{Ï€}{R}^2$

Therefore, the ratio of the dipole moment magnitudes becomes,

$\begin{array}{rcl}\frac{{\mathrm{μ}}_b}{{\mathrm{μ}}_a}& =& \frac{2i\mathrm{π}{R}_b^2}{i\mathrm{π}{R}_a^2}\\ & =& \frac{2R_b^2}{R_a^2}\end{array}$

But as the radius,

$R_b=\raisebox{1ex}{${\displaystyle R_a}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Therefore, the ratio will be,

$\begin{array}{rcl}\frac{{\mathrm{μ}}_b}{{\mathrm{μ}}_a}& =& \frac{2R_a^2}{4R_a^2}\\ & =& \frac{1}{2}\\ & =& 0.5\end{array}$

Hence, the ratio $\raisebox{1ex}{${\mathrm{μ}}_b$}\!\left/ \!\raisebox{-1ex}{${\mathrm{μ}}_a$}\right.$is $0.5$.