Question

For a toroid $N=500$, $radius=40cm$ and area of cross section $=10c{m}^2$. Find its inductance.
A. $125\mu H$
B. $250\mu H$
C. $0.00248H$
D. Zero

Answer 1

Hint: To solve the given problem we must know the formula for the magnetic field inside a toroid and also how the inductance of a coil is related to the flux through the coil and the current in the coil. Calculate the flux through the toroid and find the value of the inductance.

Formula used:
$\varphi =Li$
$B={\displaystyle \frac{{\mu }_{0}Ni}{2\pi R}}$

Complete step by step answer:
Inductance is a property of coil that relates that flux through generated by the current flowing in it.
i.e. $\varphi =Li$ …. (i),
where $\varphi$ is the flux through it, L is inductance and i is current in it.
The SI unit of inductance is Henry (H).
Toroid is a coil wound around a donut shaped ring. The coil is a conducting coil. And we know that when a current flows in a conductor, a magnetic field is produced around it.
Therefore, when a current flows in toroid, magnetic field lines are produced around it. The magnitude of the magnetic field inside a thin and small toroid is given as $B={\displaystyle \frac{{\mu }_{0}Ni}{2\pi R}}$,
where ${\mu }_0$ is a constant called permeability of free space, N is the number of turns of the coil, i is the current and R is the radius of the toroid.
Due to this magnetic field, some agentic flux is generated through the toroid.
Magnetic flux through a coil for a constant magnetic field is given as $\varphi =NBA$ …. (ii),
where A is the area of the cross section of the coil.
Substitute the value of B in (ii).
$\Rightarrow \varphi =N\left({\displaystyle \frac{{\mu }_{0}Ni}{2\pi R}}\right)A$
$\Rightarrow \varphi =\left({\displaystyle \frac{{\mu }_0{N}^{2}A}{2\pi R}}\right)i$ … (iii).
In comparing (i) and (iii) we get that $L=\left({\displaystyle \frac{{\mu }_0{N}^{2}A}{2\pi R}}\right)$.
Therefore, the inductance of a toroid is equal to $L=\left({\displaystyle \frac{{\mu }_0{N}^{2}A}{2\pi R}}\right)$ …. (iv)
It is given that $N=500$, $R=40cm=0.4m$ and $A=10c{m}^2={10}^{-3}m$.
The value of ${\mu }_0=4\pi \times {10}^{-7}Tm{A}^{-1}$
Substitute the values in (iv).
$\Rightarrow L=\left({\displaystyle \frac{4\pi \times {10}^{-7}\times {(500)}^2\times {10}^{-3}}{2\pi (0.4)}}\right)=125\times {10}^{-6}H=125\mu H$.

So, the correct answer is “Option A”.

Note:
From equation (iv), we get that the inductance of the toroid depends on the number of turns of the coils, the cross section area of the toroid and the radius of the toroid.
All the three quantities, N, A and R are physical properties of the toroid. From this we can conclude that inductance of a coil is a physical property. Meaning it only depends on the structure and design of the coil.
It is independent of the current in the coil.