Question

A thin conducting strip of width $h=2.0cm$ is tightly wound in the shape of a very long coil with cross-section radius $R=2.5cm$ to make a single-layer straight solenoid. A direct current $I=5.0A$ flows through the strip. Find the magnetic induction inside and outside the solenoid as a function of the distance $r$ from its axis.

Answer 1

If strip is tightly wound,it must have a pitch of $h$. This mean the current flow obliquely, partially along $e_z$ and $\widehat{e_{\theta }}$.

The surface current density is $\overrightarrow{F_s}$.

$F_s=\frac{1}{h}(\sqrt{1-{\left(\frac{h}{2}\pi R\right)}^2}\widehat{e_{\theta }}+\frac{h}{2\pi R}\widehat{e_z})$

By comparision with the case of a solenoid and a hollow striaght conductor,

field intensity inside the is,

$B_i={\mu }_0\frac{1}{h}1-{\left(\frac{h}{2}\pi R\right)}^2=4\times \pi \times {10}^{-7}\times \frac{5}{2\times {10}^{-5}}\sqrt{1-{\left(\frac{2}{2\pi \times 2.5}\right)}^2}=31.1\times {10}^{-5}$

magnetic field outside is due to only $\widehat{e_z}$ direction current density.

$\left(B_{out.dl}\right)=\frac{{\mu }_{0}I}{h}\times \frac{h}{2\pi R}.2\pi R{B}_{out}\times 2\pi R={\mu }_{0}I{B}_{out}=\frac{{\mu }_{0}I}{2\pi R}$