Answer

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.

**Complete step by step answer:**

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.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Which of the following distance time graph is representing class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main