Question

Figure 29-73a shows a length of wire carrying a current$\mathit{i}$and bent into a circular coil of one turn. In Fig. 29-73b the same length of wire has been bent to give a coil of two turns, each of half the original radius. (a) If${\mathit{B}}_{\mathbf{a}}$are ${\mathit{B}}_{\mathbf{b}}$the magnitudes of the magnetic fields at the centers of the two coils, what is the ratio $\raisebox{1ex}{${\mathbf{B}}_{\mathbf{b}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{B}}_{\mathbf{a}}$}\right.$? (b) What is the ratio$\raisebox{1ex}{${\mathbf{\mu }}_{\mathbf{b}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\mu }}_{\mathbf{a}}$}\right.$of the dipole moment magnitudes of the coils?

Short answer

(a) The ratio$\raisebox{1ex}{$B_b$}\!\left/ \!\raisebox{-1ex}{$B_a$}\right.$ is $4$.

(b)The ratio$\raisebox{1ex}{${\mu }_b$}\!\left/ \!\raisebox{-1ex}{${\mu }_a$}\right.$ is $0.5$.

Step-by-step solution

Listing the given quantities:

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

Understanding the concept of magnetic field:

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,

$\mu =NiA$

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

The magnetic field is,

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

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

(a) Calculations of the ratio BbBa:

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

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

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

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

The magnetic field for $a$is

$B_a=\frac{{\mu }_{0}i}{2R_a}$

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

$B_b=\frac{2{\mu }_{0}i}{2R_b}$

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

$$\begin{gathered} \frac{B_b}{B_a}=\frac{2{\mu }_{0}i}{2R_b}\times \frac{2R_a}{{\mu }_{0}i} \\ \frac{B_b}{B_a}=\frac{2R_a}{R_b} \end{gathered}$$

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$4$.

(b) Calculations of the ratio μb/μa:

The magnetic dipole moment is given as,

$\mu =NiA$

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

${\mu }_a=i{A}_a$

${\mu }_b=2i{A}_b$

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

$\frac{{\mu }_b}{{\mu }_a}=\frac{2i{A}_b}{i{A}_a}$

Since the area of circular loop is,

$A=\pi R^2$

Therefore, the ratio of the dipole moment magnitudes becomes,

$\begin{array}{rcl}\frac{{\mu }_b}{{\mu }_a}& =& \frac{2i\pi R_b^2}{i\pi R_a^2}\\ & =& \frac{2R_b^2}{R_a^2}\end{array}$

But as the radius,

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

Therefore, the ratio will be,

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

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