Answer

Verified

98.1k+ 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

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main