Chapter 29, Problem 38AP

Two circular coils of radius R, each with N turns, are perpendicular to a common axis. The coil centers are a distance R apart. Each coil carries a steady current I in the same direction as shown in Figure P29.38. (a) Show that the magnetic field on the axis at a distance x from the center of one coil is

$B=\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{(R^2+x^2)}^{3/2}}+\frac{1}{{(2R^2+x^2-2Rx)}^{3/2}}\right]$

To determine

To show: The magnetic field on the axis at a distance $x$ from the center of one coil is $B=\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{\left(R^2+x^2\right)}^{3/2}}+\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{3/2}}\right]$.

Answer to Problem 38AP

Hence, the magnetic field on the axis at a distance $x$ from the center of one coil is $B=\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{\left(R^2+x^2\right)}^{3/2}}+\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{3/2}}\right]$.

Explanation of Solution

Given information: The radius of each coil is $R$, number of turns is $N$, distance between two coils is $R$ and the value of steady current is $I$.

Let consider a point at $x$ distance from one ring then the distance of this point from the second ring will be $\left(R-x\right)$.

Formula to calculate the magnetic field at a point due to first circular coil is,

$B_1=\frac{{\mu }_{0}NI{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}$ (1)

Here,

$B_1$ is the magnetic field on the axis of first circular ring.

${\mu }_0$ is the absolute permittivity.

$I$ is the current flows in the circular ring.

$x$ is the distance of the of point from the center of the ring.

$R$ is the radius of the circular ring.

$N$ is the number of turns in each coil.

Formula to calculate the magnetic field to the same point due to second circular coil is,

$B_2=\frac{{\mu }_{0}NI{R}^2}{2{\left({\left(R-x\right)}^2+R^2\right)}^{3/2}}$ (2)

Here,

$B_2$ is the magnetic field on the axis of second circular ring.

Write the expression for the net field at the given point.

$B=B_1+B_2$

Substitute $\frac{{\mu }_{0}NI{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}$ for $B_1$ and $\frac{{\mu }_{0}NI{R}^2}{2{\left({\left(R-x\right)}^2+R^2\right)}^{3/2}}$ for $B_2$ in above equation.

$B=\frac{{\mu }_{0}NI{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}+\frac{{\mu }_{0}NI{R}^2}{2{\left({\left(R-x\right)}^2+R^2\right)}^{3/2}}$

Simplify the above equation for $B$.

$\begin{array}{c}B=\frac{{\mu }_{0}NI{R}^2}{2}\left[\frac{1}{{\left(x^2+R^2\right)}^{3/2}}+\frac{1}{{\left({\left(R-x\right)}^2+R^2\right)}^{3/2}}\right]\\ =\frac{{\mu }_{0}NI{R}^2}{2}\left[\frac{1}{{\left(x^2+R^2\right)}^{3/2}}+\frac{1}{{\left(R^2+x^2-2Rx+R^2\right)}^{3/2}}\right]\\ =\frac{{\mu }_{0}NI{R}^2}{2}\left[\frac{1}{{\left(x^2+R^2\right)}^{3/2}}+\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{3/2}}\right]\end{array}$ (3)

Conclusion:

Therefore, the magnetic field on the axis at a distance $x$ from the center of one coil is $B=\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{\left(R^2+x^2\right)}^{3/2}}+\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{3/2}}\right]$.

To determine

To show: The value of $dB/dx$ and $d{B}^2/d{x}^2$ are both zero at the point midway between the coils.

Answer to Problem 38AP

Hence, the value of $dB/dx$ and $d{B}^2/d{x}^2$ are both zero at the point midway between the coils.

Explanation of Solution

Given information: The radius of each coil is $R$, number of turns is $N$, distance between two coils is $R$ and the value of steady current is $I$.

From equation (3), the expression for the net magnetic field is given by,

$B=\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{\left(R^2+x^2\right)}^{3/2}}+\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{3/2}}\right]$

Differentiate the above equation with respect to $x$.

$\begin{array}{c}\frac{dB}{dx}=\frac{d\left(\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{\left(R^2+x^2\right)}^{3/2}}+\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{3/2}}\right]\right)}{dx}\\ =\frac{N{\mu }_{0}I{R}^2}{2}\left(\frac{d\left({\left(R^2+x^2\right)}^{-3/2}\right)}{dx}+\frac{d\left({\left(2R^2+x^2-2Rx\right)}^{-3/2}\right)}{dx}\right)\\ =\frac{N{\mu }_{0}I{R}^2}{2}\left[\begin{array}{l}\left(\frac{-3}{2}\right)\left(\frac{1}{{\left(R^2+x^2\right)}^{5/2}}\right)\left(2x\right)+\\ \left(\frac{-3}{2}\right)\left(\frac{1}{{\left(2R^2+x^2-2Rx\right)}^{5/2}}\right)\left(2x-2R\right)\end{array}\right]\\ =\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\left(\frac{x}{{\left(R^2+x^2\right)}^{5/2}}\right)+\left(\frac{x-R}{{\left(2R^2+x^2-2Rx\right)}^{5/2}}\right)\right]\end{array}$ (4)

Substitute $\frac{R}{2}$ for $x$ in the above equation.

$\begin{array}{c}\frac{dB}{dx}=\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\left(\frac{\frac{R}{2}}{{\left(R^2+{\left(\frac{R}{2}\right)}^2\right)}^{5/2}}\right)+\left(\frac{\frac{R}{2}-R}{{\left(2R^2+{\left(\frac{R}{2}\right)}^2-2R\left(\frac{R}{2}\right)\right)}^{5/2}}\right)\right]\\ =\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\left(\frac{\frac{R}{2}}{{\left({\frac{5R}{4}}^2\right)}^{5/2}}\right)+\left(\frac{-\frac{R}{2}}{{\left({\frac{5R}{4}}^2\right)}^{5/2}}\right)\right]\\ =\frac{-3N{\mu }_{0}I{R}^2}{2}\times 0\\ =0\end{array}$

Thus, the value of value of $dB/dx$ is zero at the point midway between the coils.

Differentiate the equation (4) with respect to $x$.

$\begin{array}{c}\frac{d{B}^2}{d{x}^2}=\frac{d\left(\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\left(\frac{x}{{\left(R^2+x^2\right)}^{5/2}}\right)+\left(\frac{x-R}{{\left(2R^2+x^2-2Rx\right)}^{5/2}}\right)\right]\right)}{dx}\\ =\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\frac{d\left(\frac{x}{{\left(R^2+x^2\right)}^{5/2}}\right)}{dx}+\frac{d\left(\frac{x-R}{{\left(2R^2+x^2-2Rx\right)}^{5/2}}\right)}{dx}\right]\\ =\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\begin{array}{l}\frac{1}{{\left(R^2+x^2\right)}^{5/2}}-\frac{5x^2}{{\left(R^2+x^2\right)}^{7/2}}+\\ \frac{1}{{\left(2R^2+x^2-2Rx\right)}^{5/2}}-\frac{5{\left(x-R\right)}^2}{{\left(2R^2+x^2-2Rx\right)}^{7/2}}\end{array}\right]\end{array}$

Substitute $\frac{R}{2}$ for $x$ in the above equation.

$\begin{array}{l}\frac{d{B}^2}{d{x}^2}=\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\begin{array}{l}\frac{1}{{\left(R^2+{\left(\frac{R}{2}\right)}^2\right)}^{5/2}}-\frac{5{\left(\frac{R}{2}\right)}^2}{{\left(R^2+{\left(\frac{R}{2}\right)}^2\right)}^{7/2}}+\\ -\frac{1}{{\left(2R^2+{\left(\frac{R}{2}\right)}^2-2R\left(\frac{R}{2}\right)\right)}^{5/2}}-\frac{5{\left(\left(\frac{R}{2}\right)-R\right)}^2}{{\left(2R^2+{\left(\frac{R}{2}\right)}^2-2R\left(\frac{R}{2}\right)\right)}^{7/2}}\end{array}\right]\\ =\frac{-3N{\mu }_{0}I{R}^2}{2}\left[\begin{array}{l}\frac{1}{{\left(\frac{5R^2}{4}\right)}^{5/2}}-\frac{5{\left(\frac{R}{2}\right)}^2}{{\left(\frac{5R^2}{4}\right)}^{7/2}}+\\ -\frac{1}{{\left(\frac{5R^2}{4}\right)}^{5/2}}-\frac{-5{\left(\frac{R}{2}\right)}^2}{{\left(\frac{5R^2}{4}\right)}^{7/2}}\end{array}\right]\\ =0\end{array}$

Thus, the value of value of $\frac{d{B}^2}{d{x}^2}$ is zero at the point midway between the coils.

Conclusion:

Therefore, the value of $dB/dx$ and $d{B}^2/d{x}^2$ are both zero at the point midway between the coils.