Question

(a) Using Biot-Savart's law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop.
(b) What does a toroid consist of ? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero.

Answer 1

(a) Magnetic field on the axis of a circular loop
I $\to$ Current
R $\to$ Radii
X $\to$ Axis
x $\to$ Distance of OP
dl $\to$ Conducting element of the loop
According to Biot-Savart's law, the magnetic field at P is

According to Biot-Savart's law
$dB=\frac{{\mu }_0}{4\pi }\frac{I\overrightarrow{d}l\times \overrightarrow{r}}{r^2}$
$dB=\frac{{\mu }_0}{4\pi }\frac{Idlsin\theta }{r^2}$
$dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{r^2}[\because \overrightarrow{d}l\perp \hat{r}]$
dB has two components $d{B}_x$ and $d{B}_y$, $d{B}_y$ components being in opposite direction cancel out each other and $d{B}_x$ component being in same direction added up.

$\therefore d{B}_x=dBsin\theta$

$=\frac{{\mu }_{0}Idl}{4\pi r^2}\times \frac{R}{(x^2+R^2{)}^{\frac{1}{2}}}$

$d{B}_x=\frac{{\mu }_{0}Idl}{4\pi }\times \frac{R}{(x^2+R^2{)}^{\frac{1}{2}}}$
$[\because r^2=x^2+R^2]$

Total magnetic field at P

$B=\int d{B}_x$

$=\int \frac{{\mu }_{0}IRdl}{4\pi (x^2+R^2{)}^{\frac{3}{2}}}$

$\frac{{\mu }_{0}IR}{4\pi (x^2+R^2{)}^{\frac{3}{2}}}\int dl=\frac{{\mu }_{0}IR}{4\pi (x^2+R^2{)}^{\frac{3}{2}}}\times 2\pi RB=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{\frac{3}{2}}}\underset{B}{\to }=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{\frac{3}{2}}}\hat{i}$
(b) Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound.
Figure shows a sectional view of the toroid. The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines.

By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop.
Let the magnetic field inside the toroid be B. Then by Ampere's circuital law,
$\int \overrightarrow{B}.\overrightarrow{dl}={\mu }_{0}I\Rightarrow BL={\mu }_{0}NI$

Where, L is the length of the loop for which B is tangential, I is the current enclosed by one loop and N is the number of turns.
We find, $L=2\pi r$
The current threads the ring as many times are there are turns in the solenoid, therefore total current in the solenoid is NI.
$B\left(2\pi r\right)={\mu }_{0}NI$
or $B=\frac{{\mu }_{0}NI}{2\pi r}$
Open space inside the toroid encloses no current thus, I = 0.
Hence, B = 0
Open space exterior to the toroid :
Each turn of current carrying wire is cut twice by the loop 3. Thus, the current coming out of the plane of the paper is cancelled exactly by the current going into it. Thus, I= 0, and B = 0