Question

A thin insulated wire forms a plane spiral of $N=100$ tight turns carrying a current $I=8mA$. The radii of inside and outside turns (Fig.) are equal to $a=50mm$ and $b=100mm$ Find:
(a) the magnetic induction at the centre of the spiral;
(b) the magnetic moment of the spiral with a given current.

Answer 1

(a) From Biot-Savart's law, the magnetic induction due to a circular current carrying wire loop at its centre is given by,
$B_r=\frac{{\mu }_0}{2r}i$
The plane spiral is made up of concentric circular loops, having different radii, varying from $a$ to $b$. Therefore, the total magnetic induction at the centre,
$B_0=\int \frac{{\mu }_0}{2r}dN....\left(1\right)$
Where $\frac{{\mu }_0}{2}i$ is the contribution of one turn of radius $r$ and $dN$ is the number of turns in the interval $(r,r+dr)$
i.e. $dN=\frac{N}{b-a}dr$
Substituting in equation (1) and integrating the result over $r$ between $a$ and $b$, we obtain,
$B_0={\int }_a^b\frac{{\mu }_{0}i}{2r}\frac{N}{(b-a)}dr=\frac{{\mu }_{0}iN}{2(b-a)}ln\frac{b}{a}$
(b) The magnetic moment of a turn of radius $r$ is $p_m=i\pi r^2$ and of all turns,
$p=\int p_{m}dN={\int }_a^{b}i\pi r^2\frac{N}{b-a}dr=\frac{\pi iN(b^3-a^2)}{3(b-a)}$.