Question

Use the Biot-Savart law (most conveniently in the form of Eq. 5.42 appropriate to surface currents) to find the field inside and outside an infinitely long solenoid of radiusR, with n turns per unit length, carrying a steady current I.

Short answer

The value of magnetic field inside the solenoid is$\overrightarrow{B}={\mu }_{0}nI\stackrel{⏜}{z}$

The value of magnetic field outside the solenoid is zero.

Step-by-step solution

Write the given data from the question.

Consider the given an infinitely long solenoid of radiusR, with n turns per unit length, carrying a steady current I.

Determine the formula of magnetic field outside the solenoid.

Write the formula of magnetic field for outside and inside the solenoid.

${\mathbf{B}}_{\mathbf{z}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{k}}{\mathbf{2}}\left[\frac{R^2-s^2}{\left|R^2-s^2\right|}+1\right]$ …… (1)

Here, ${\mathit{\mu }}_{\mathbf{0}}$is permeability, $\mathbf{R}$is solenoid radiusR,$\mathbf{s}$ is radius inside the solenoid.

Determine the value of magnetic field outside and inside the solenoid.

The magnetic field created in a wire at any location as a result of current is determined by summing the contributions to the magnetic field made by tiny wire segments under Biot- Savart's law.

Draw a circuit diagram of solenoid in Cartesian space.

Assume field point on the x-axis.

So,

$r=\left(s,0,0\right)$

Then,

Magnetic field is given by,

role="math" localid="1657689928279" $$\begin{gathered} B=\frac{{\mu }_0}{4\mathrm{\pi }}\int \left(\frac{k\times r}{r^2}\right)da \\ da=Rd\varphi dz \\ K=k\stackrel{⏜}{\varphi } \\ \stackrel{⏜}{\varphi }=\left(-\sin \varphi \stackrel{⏜}{x}+\cos \varphi \stackrel{⏜}{y}\right) \end{gathered}$$

Substitute value of into above equation.

$K=k\left(-\sin \varphi \stackrel{⏜}{x}+\cos \varphi \stackrel{⏜}{y}\right)$

Solve as:

$$\begin{gathered} r=\left(s-R\cos \varphi \right)\stackrel{⏜}{x}-R\sin \varphi \stackrel{⏜}{y}-z\stackrel{⏜}{z} \\ K\times r=k\left|\begin{array}{ccc}\stackrel{⏜}{x}& \stackrel{⏜}{y}& \stackrel{⏜}{z}\\ -\sin \varphi & \cos \varphi & 0\\ s-R\cos \varphi & \cos \varphi & -z\end{array}\right| \\ K\times r=k\left[\left(-2\cos \varphi \right)\stackrel{⏜}{x}+\left(-z\sin \varphi \right)\stackrel{⏜}{y}\right]+\left(R-s\cos \varphi \right)\stackrel{⏜}{z} \\ r=z^2+R^2+s^2-2Rs\cos \varphi \end{gathered}$$

Determine X and Y components of magnetic field,

$$\begin{gathered} B_2=\frac{{\mu }_0}{4\mathrm{\pi }}KR\int \frac{\left(R-\cos \varphi \right)}{{\left(z^2+R^2+s^2-32Rs\cos \varphi \right)}^{{\displaystyle \frac{3}{2}}}}d\varphi dz \\ {B}_2=\frac{{\mu }_{0}KR}{4\mathrm{\pi }}\int \left(R-s\cos \varphi \right)\int \frac{dz}{{\left(z^2+d^2\right)}^{{\displaystyle \frac{3}{2}}}}d\varphi \end{gathered}$$

Here,$d=R^2+s^2-2Rs\cos \varphi$. Solve as:

$$\begin{gathered} {\int }_{-\infty }^{+\infty }\frac{dz}{{\left(a^2+d^2\right)}^{{\displaystyle \frac{3}{2}}}}={\left[\frac{2z}{d^2\sqrt{z^2+d^2}}\right]}_0^{\infty } \\ =\frac{2}{d^2} \end{gathered}$$

Further solve,

$$\begin{gathered} B_z=\frac{{\mu }_{0}KR}{2\mathrm{\pi }}\int \frac{\left(R-s\cos \varphi \right)}{\left(R^2+s^2-2Rs\cos \varphi \right)}d\varphi \\ R=s\cos \varphi =\frac{1}{2R}\left[\left(R^2-s^2\right)+\left(R^2+s^2\right)-2R\cos \varphi \right] \\ {B}_z=\frac{{\mu }_{0}KR}{2\mathrm{\pi }\left(2\mathrm{R}\right)}{\int }_0^{2\mathrm{\pi }}\frac{\left(\left(R^2-s^2\right)+R^2+s^2-2Rs\cos \varphi \right)d\varphi }{R^2+s^2-2Rs\cos \varphi } \\ {B}_z=\frac{{\mu }_{0}K}{4\mathrm{\pi }}\left[R^2-s^2{\int }_0^{2\mathrm{\pi }}\frac{d\varphi }{R^2+s^2-2Rs\cos \varphi }+{\int }_0^{2\mathrm{\pi }}d\varphi \right] \end{gathered}$$

Use properly to evaluate ${\int }_0^{2\mathrm{\pi }}\frac{d\varphi }{R^2+s^2-2Rs\cos \varphi }$,

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}\frac{d\varphi }{a+b\cos \varphi }=2{\int }_0^{\mathrm{\pi }}\frac{d\varphi }{a+b\cos \varphi } \\ =\frac{4}{\sqrt{a^2-b^2}}{\tan }^{-1}{\left[\frac{\sqrt{a^2{b}^2+\tan {\displaystyle \frac{\varphi }{2}}}}{a+b}\right]}_0^{\mathrm{\pi }} \end{gathered}$$

Consider the expressions:

$$\begin{gathered} a=R^2+a^2 \\ b=-2Rs \end{gathered}$$

Solve as:

$$\begin{gathered} {\int }_0^2\frac{d\varphi }{a+b\cos \varphi }=\frac{4}{\sqrt{a^2+b^2}}{\tan }^{-1}\left[\frac{\sqrt{a^2+b^2}\tan {\displaystyle \frac{\mathrm{\pi }}{2}}}{a+b}\right] \\ =\frac{4}{\sqrt{a^2+b^2}}{\tan }^{-1}\left(\infty \infty \right) \\ =\left(\frac{4}{\sqrt{a^2+b^2}}\right)\frac{\mathrm{\pi }}{2} \\ =\frac{2}{\sqrt{a^2-b^2}} \end{gathered}$$

Now subtract the $\left(a-b\right)$.

$$\begin{gathered} a^2-b^2={\left(R^2+s^2\right)}^2-{\left(-2Rs\right)}^2 \\ =R^4+2R^2{s}^2+s^4-4R^2{s}^2 \\ =R^4-2R^2{s}^2+s^4 \\ ={\left(R^2-s^2\right)}^2 \end{gathered}$$

Solve further as

role="math" localid="1657691894443" $\sqrt{a^2-b^2}=\left|R^2-s^2\right|$

Substitute $\left|R^2-s^2\right|$for$\sqrt{a^2-b^2}$into above equation.

$$\begin{gathered} B_z=\frac{{\mu }_{0}K}{4\mathrm{\pi }}\left[\frac{\left(R^2-s^2\right)}{\left|R^2-s^2\right|}2\mathrm{\pi }+2\mathrm{\pi }\right] \\ =\frac{{\mu }_{0}K\left(2\mathrm{\pi }\right)}{4\mathrm{\pi }}\left[\frac{R^2-s^2}{\left|R^2-s^2\right|}+1\right] \\ =\frac{{\mu }_{0}K}{2}\left[\frac{R^2-s^2}{\left|R^2-s^2\right|}+1\right] \end{gathered}$$

Determine the magnetic field inside the solenoid.

$s<R$

Substitute $1for\left(\frac{R^2-s^2}{\left|R^2-s^2\right|}\right)$into equation (1)

role="math" localid="1657693115344" $$\begin{gathered} B_z=\frac{{\mu }_{0}k}{2}\left(1+1\right) \\ ={\mu }_{0}k \end{gathered}$$

Substitute $K=nI$into above equation.

$B_z={\mu }_{0}nI$

Determine the magnetic field outside the solenoid.

$s>R$

Substitute -1 for$\left(\frac{R^2-s^2}{\left|R^2-s^2\right|}\right)$ into equation (1)

$$\begin{gathered} B_z=\frac{{\mu }_{0}k}{2}\left(-1+1\right) \\ =0 \end{gathered}$$

Therefore, the value of magnetic field inside the solenoid is $\overrightarrow{B}={\mu }_{0}nI\stackrel{⏜}{z}$and outside

the solenoid is zero.