Show that the inductance of a toroid of rectangular cross-section is given by $$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$ where (N) is the total number of turns, (a) is the inner radius, (b) is the outside radius and (H) is the height of the toroid.
Solution:
Toroid
Using the definition of the self inductance of a solenoid, we express (L) in terms of flux (\phi), (N) and (I):
$$ L=\frac{N\phi}{I}$$
We apply Ampere's law to a closed path of radius (a<r<b):
$$ \oint \vec{B}.\vec{dl}=B(2\pi r)$$ $$=\mu_0NI$$ $$ \Rightarrow B=\frac{\mu_0NI}{2\pi r}$$ We express the flux in a strip of height (H) and width (dr):
$$ d\phi=BHdr=\frac{\mu_0NIH}{2\pi}\int_{a}^{b}\frac{dr}{r}$$ $$ =\frac{\mu_0NIH}{2\pi}ln(\frac{b}{a})$$
Substitute for flux (\phi) in the equation (1) we obtain the expression for (L)
$$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$
Magnetic Field at the Centre of a Ring
A ring of radius (R) carries a linear charge density ($\lambda$). It is rotating with angular speed ($\omega$). What is the magnetic field at the centre?
Discussion:
Linear charge density $$ \lambda=\frac{Q}{2\pi R}
$$
When the ring is rotated about the axis, the motion of the electrons in a circular orbit is equivalent to a current carrying loop.
Current $$ I=\frac{Q}{T}=\frac{Q\omega}{2\pi}$$
since Time period (T=2\pi/\omega).
Now, magnetic field around the centre of a current carrying loop is given by $$ B=\mu_0I/2R$$
Putting the value of (I) in the above equation, we get
$$ B=\frac{\mu_0\omega}{2}.\frac{Q}{2\pi R}
$$$$ \Rightarrow B=\frac{\mu_0\lambda\omega}{2}
$$
Magnetic Field at Focus of Parabola
Let's solve the problem based on Magnetic Field at Focus of Parabola and learn how to solve it. First, try it yourself, then check your solution.
The Problem:
An infinite wire carrying current (I) is bent in the form of a parabola. Find the magnetic field at the focus of the parabola. Take the distance of the focus from the apex as (a).
Solution:
From Biot-Savart law, the magnetic field ar (S) is given by $$ \vec{B}= \frac{\mu_0}{4 \pi} \int\frac{I\vec{dl}\times\vec{r}}{r^3}$$ From the figure, we note that $$ |\vec{dl}\times \vec{r}|$$=area of the parallelogram by (\vec{dl}) and (\vec{r}) $$ = 2\times1/2\times r.rd\theta$$$$=r^2d\theta$$ Hence, $$ \vec{B}=\frac{\mu_0I}{4 \pi}\int_{0}^{2\pi}\frac{d\theta}{r}$$ Using (r(1-cos\theta)=2a) as the equation to the parabola, we get $$ \vec{B}=\mu_0I/4a $$
