Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A toroid has a mean radius $R$ equal to $\dfrac{{20}}{\pi }cm$, and a total of $400$ turns of wire carrying a current of $2.0A$. An aluminium ring at a temperature $280K$ inside the toroid provides the core. If the magnetization $I$ is $4.8 \times {10^{ - 2}}A{m^{ - 1}}$, the susceptibility of aluminium at $280K$ is.

Last updated date: 20th Jun 2024
Total views: 53.7k
Views today: 0.53k
Verified
53.7k+ views
Hint: First we need to calculate the number of turns of the wire per unit length. Then we can find the magnetic intensity by multiplying the number of turns per unit length with the current in the wire. The susceptibility will be given by magnetization divided by the magnetic intensity.

Formula Used In this solution we will be using the following formula,
$n = \dfrac{N}{L}$
where $n$ is the number of turns per unit length
$N$ is the total number of turns, and $L$ is the length of the toroid.
$H = ni$
where $H$ is the magnetic intensity and $i$ is the current
and $\chi = \dfrac{I}{H}$
where $\chi$ is the susceptibility and $I$ is the magnetization.

In the question we are given a toroid of radius of $\dfrac{{20}}{\pi }cm$. So the whole length of the toroid will be the perimeter of this circle.
So the length will be, $L = 2\pi \times \dfrac{{20}}{\pi }$
Cancelling the $\pi$ we get, $L = 2 \times 20 = 40cm$, that is, $L = 40 \times {10^{ - 2}}m$
Now the total number of turns on the toroid is given by, $N = 400$
So the number of turns per unit length will be the total number of turns by the length. This is given as,
$n = \dfrac{N}{L}$
Now substituting the values we get,
$\Rightarrow n = \dfrac{{400}}{{40 \times {{10}^{ - 2}}}}$
This gives us on calculating,
$\Rightarrow n = 10 \times {10^{ - 2}}/m$
The magnetic intensity inside the core will be $H = ni$
In the question we are given the current in the wire as, $i = 2.0A$
So substituting the value of $n$ and $i$ in the formula for the magnetic intensity we get,
$\Rightarrow H = 10 \times {10^{ - 2}} \times 2.0$
So on calculating we get,
$\Rightarrow H = 2 \times {10^{ - 2}}A/m$
The magnetization of aluminium is given as $I = 4.8 \times {10^{ - 2}}A{m^{ - 1}}$. So using this magnetization and magnetic intensity, we can find the susceptibility of the material by the formula,
$\chi = \dfrac{I}{H}$
Substituting we get,
$\Rightarrow \chi = \dfrac{{4.8 \times {{10}^{ - 2}}A/m}}{{2 \times {{10}^{ - 2}}A/m}}$
On calculating we get,
$\Rightarrow \chi = \dfrac{{4.8}}{2}$

Therefore, the susceptibility is given as, $\chi = 2.4$.

Note: The susceptibility of a material is the amount by which a material will become magnetised when it is placed in a magnetic field. It is given by the ratio of magnetisation and magnetic field intensity. The materials having susceptibility more than 0 are paramagnetic and the ones having susceptibility less than 0 are diamagnetic.