91Ó°ÊÓ

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

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, ${\mathrm{Ï•}}_B=Li$

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

3. Inductance of the coil with N turns,

$L=\frac{N{\mathrm{Ï•}}_B}{i}$

Magnetic flux over the surface attached to the closed loop

${\mathrm{Ï•}}_B=âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{A}=BA=2.60×{10}^{-3}×\mathrm{Ï€}×{\left(10×{10}^{-2}\right)}^2=8.164×{10}^{-7}\mathrm{Wb}$

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

Magnetic flux through the coil having N turns

$$\begin{gathered} \mathrm{Ï•}={\mathrm{\pm ·Ï•}}_{\mathrm{B}}=30×8.164×{10}^{-7}=2.449×{10}^{-3}\mathrm{Wb} \\ \mathrm{Ï•}≈2.45×{10}^{-3}\mathrm{Wb} \end{gathered}$$

Inductance of the coil with N turns

$$\begin{gathered} L=\frac{N{\mathrm{ϕ}}_B}{i} \\ =\frac{\mathrm{ϕ}}{i} \\ L=6.447×{10}^{-4}\mathrm{H} \\ ≈6.45×{10}^{-4}\mathrm{H} \end{gathered}$$

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