Figure shows a copper strip of width W = 16.0 cmthat has been bent to form a shape that consists of a tube of radius R = 1.8 cmplus two parallel flat extensions. Current i = 35 mAis distributed uniformly across the width so that the tube is effectively a one-turn solenoid. Assume that the magnetic field outside the tube is negligible and the field inside the tube is uniform. (a) What is the magnetic field magnitude inside the tube? (b) What is the inductance of the tube (excluding the flat extensions)?

1. Width of the copper strip, $W=16.0\mathrm{cm}=16.0\times {10}^{-2}\mathrm{m}$
2. Radius of the copper tube ,$R=1.8\times {10}^{-2}\mathrm{m}$
3. Current distributed across the width of the tube ,$i=35.0\mathrm{mA}=35.0\times {10}^{-3}\mathrm{A}$

For this problem, we can approximate the copper tube to be made up of closely spaced circular loops. So it forms a solenoid with Nturns of radiusand the current is uniformly distributed over all these circular loops or solenoids with Nturns. By this approximation, one can find the field inside the tube.

Formula:

1. Magnetic flux ${\varphi }_B=Li=\oint \overrightarrow{B}.d\overrightarrow{A}$
2. Number of turns per unit length of the solenoid,$n=\frac{N}{L}$
3. Magnetic field inside solenoid ,$B={\mu }_{0}nl$
4. Inductance of coil,

$L=\frac{{\varphi }_B}{i}$

Let’s assume that the current flowing through each circular loop to be $\delta i$ , we know that total

Current through the tube is i and this is distributed uniformly over each loop so we get

$$\begin{gathered} \delta i\times N=i \\ \delta i=\frac{i}{N} \end{gathered}$$

Now Number of turns per unit width of the solenoid, $n=\frac{N}{W}$and

Magnetic field inside solenoid,

$B={\mu }_{0}nl$

Using this two relations we get

$$\begin{gathered} B={\mu }_{0}nl={\mu }_0.\frac{N}{W}.\delta i \\ B={\mu }_0.\frac{N}{W}.\frac{i}{N} \\ B=\frac{{\mu }_{0}i}{W} \\ B=\frac{4{\mathrm{\pi \mu }}_0\mathrm{i}}{4\mathrm{\pi W}} \\ B=\frac{4\mathrm{\pi }\times {10}^{-7}\times 35.0\times {10}^{-3}}{16.0\times {10}^{-2}} \\ B=27.47\times {10}^{-8} \\ B=2.74\times {10}^{-7}T \end{gathered}$$

Magnetic flux,

$$\begin{gathered} {\varphi }_B=\oint \overrightarrow{B}.d\overrightarrow{A} \\ B=BA \\ B=B{\mathrm{\pi R}}^2 \\ {\mathrm{\varphi }}_{\mathrm{B}}={\mathrm{B\pi R}}^2 \\ {\mathrm{\varphi }}_{\mathrm{B}}=2.74\times {10}^{-7}\times \mathrm{\pi }\times {\left(1.8\times {10}^{-2}\right)}^2 \\ {\mathrm{\varphi }}_{\mathrm{B}}=2.787\times {10}^{-10}\mathrm{Wb} \end{gathered}$$

Inductance of coil,

$$\begin{gathered} L=\frac{{\varphi }_B}{i} \\ L=\frac{2.787\times {10}^{-10}}{35.0\times {10}^{-3}} \\ L=0.0796\times {10}^{-7}\approx 8.0\times {10}^{-9}\mathrm{H} \end{gathered}$$