Question

A circular coil has a 10.0 cm radius and consists of 30.0closely wound turns of wire. An externally produced magnetic field of magnitude 2.60mTis perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns? (b) When the current in the coil is 3.80 Ain a certain direction, the net flux through the coil is found to vanish. What is the inductance of the coil?

Short answer

$$\begin{gathered} a)\mathrm{\varphi }\approx 2.45\times {10}^{-3}\mathrm{Wb} \\ b)L=6.447\times {10}^{-4}\mathrm{H}\approx 6.45\times {10}^{-4}\mathrm{H} \end{gathered}$$

Step-by-step solution

Given

1. Radius of the circular coil,$R=10.0\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$
2. Number of turns, N = 30
3. Magnetic field strength, $B_{ext}=2.60\times {10}^{-3}\mathrm{T}$
4. Current though the coil, i = 3.80 A

Understanding the concept

We know that magnetic flux through the coil is proportional to the current flowing through that coil and the proportionality constant is the inductance of that coil. The inductance of the coil depends on the number of turns of the coil. Flux through a single turn multiplied by a total number of turns gives the total flux through that coil.

Formula:

1. Magnetic flux, ${\varphi }_B=Li$

2. Magnetic flux through the coil having N turns, $\varphi =N{\varphi }_B$

3. Inductance of the coil with N turns,

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

4.${\varphi }_B=\oint \overrightarrow{B}.d\overrightarrow{A}$, magnetic flux over the surface attached to the closed loop

(a) Calculate the magnetic flux linked to its turns if there is no current is in the coil

Magnetic flux over the surface attached to the closed loop

${\varphi }_B=\oint \overrightarrow{B}.d\overrightarrow{A}=BA=2.60\times {10}^{-3}\times \mathrm{\pi }\times {\left(10\times {10}^{-2}\right)}^2=8.164\times {10}^{-7}\mathrm{Wb}$

Where $A=\mathrm{\pi }{R}^2$is the area of the loop now

Magnetic flux through the coil having N turns

$$\begin{gathered} \mathrm{\varphi }={\mathrm{N\varphi }}_{\mathrm{B}}=30\times 8.164\times {10}^{-7}=2.449\times {10}^{-3}\mathrm{Wb} \\ \mathrm{\varphi }\approx 2.45\times {10}^{-3}\mathrm{Wb} \end{gathered}$$

(b) Calculate the inductance of the coil

Inductance of the coil with N turns

$$\begin{gathered} L=\frac{N{\varphi }_B}{i} \\ =\frac{\varphi }{i} \\ L=6.447\times {10}^{-4}\mathrm{H} \\ \approx 6.45\times {10}^{-4}\mathrm{H} \end{gathered}$$

Therefore, the inductance of the coil is $6.45\times {10}^{-4}\mathrm{H}$.