Chapter 30, Problem 60AP

(a)

To determine

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

Answer to Problem 60AP

The proof for the statement that the magnetic field at a distance $x$ from center of one coil is $\frac{N{\mu }_{0}I{R}^2}{2}\left[\frac{1}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}+\frac{1}{{\left(x^2+2R^2-2Rx\right)}^{\frac{3}{2}}}\right]\widehat{i}$ is stated as below.

Explanation of Solution

The following figure shows two circular coils of radius $R$ that are perpendicular to a common axis.

Figure-(1)

From Figure (1) consider that the distance $x$ from point $P$ to the center of the left coil.

Write the expression for the magnetic field at point $P$ due to left coil.

    $\overrightarrow{B_1}=\frac{N{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}\widehat{i}$

Here, $\overrightarrow{B_1}$ is the magnetic field at point $P$ due to left coil, $N$ is the number of turns on the coil, ${\mu }_0$ is the permeability of free space, $R$ is the radius of the coil and $x$ is the distance between the center of the left coil and point $P$.

Write the expression for the magnetic field at point $P$ due to right coil.

    $\overrightarrow{B_2}=\frac{N{\mu }_{0}I{R}^2}{2{\left({\left(x-R\right)}^2+R^2\right)}^{\frac{3}{2}}}\widehat{i}$

Here, $\overrightarrow{B_2}$ is the magnetic field at point $P$ due to the right coil.

From Figure (1) it is shown that the current is flowing on the same direction. Therefore the expression for the net magnetic field at point $P$ due to both the coils are.

    $B=\overrightarrow{B_1}+\overrightarrow{B_2}$                                                                                                                (I)

Here, $B$ is the net magnetic field at point $P$ due to both the coils.

Substitute $\frac{N{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{\frac{3}{2}}}\widehat{i}$ for $\overrightarrow{B_1}$ and $\frac{N{\mu }_{0}I{R}^2}{2{\left({\left(x-R\right)}^2+R^2\right)}^{\frac{3}{2}}}\widehat{i}$ for $\overrightarrow{B_2}$ in equation (I) to calculate $B$.

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

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

(b)

To determine

The proof that $\frac{dB}{dx}$ and $\frac{d^{2}B}{d{x}^2}$ are equal to zero at the point midway between the coils.

Answer to Problem 60AP

The proof that $\frac{dB}{dx}$ and $\frac{d^{2}B}{d{x}^2}$ are equal to zero at the point midway between the coils is as stated below.

Explanation of Solution

The magnetic field at the point midway between the coils is,

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

Differentiate the above equation with respect to $x$.

    $\frac{dB}{dx}=\frac{N{\mu }_{0}I{R}^2}{2}\left[\left(\frac{-3}{2}\right)\left(\frac{2x}{{\left(x^2+R^2\right)}^{\frac{5}{2}}}+\frac{2\left(R-x\right)\left(-1\right)}{{\left({\left(R-x\right)}^2+R^2\right)}^{\frac{5}{2}}}\right)\right]$                                        (II)

Substitute $\frac{R}{2}$ for $x$ in Equation (I) to calculate $\frac{dB}{dx}$.

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

Differentiate Equation (II) with respect to $x$ to calculate $\frac{d^{2}B}{d{x}^2}$.

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

Substitute $\frac{R}{2}$ for $x$ in the above to calculate $\frac{d^{2}B}{d{x}^2}$.

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

Further solve the above equation.

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

Conclusion:

From Equation (III) and Equation (IV), $\frac{dB}{dx}$ and $\frac{d^{2}B}{d{x}^2}$ are equal to zero. Therefore, magnetic fields at the midway between the coils are uniform.

Therefore, both the coils are Helmholtz coils.