Question

Aluminum has a magnetic susceptibility χ = 1.7 × 10-5 at T = 300 K. What will be the magnetization of a small sample of aluminum placed in a 1.5 T magnetic field at T = 150 K?

Answer 1

Answer:

The magnetization of a small sample of aluminum is 40.6 A/m

Explanation:

Given that,

Magnetic susceptibility $\chi_{1}=1.7\times10^{-5}$ at 300 K

Magnetic field = 1.5 T

We need to calculate the magnetic susceptibility at 150 K

$\dfrac{\chi_{2}}{\chi_{1}}=\dfrac{T_{1}}{T_{2}}$

Put the value into the formula

$\dfrac{\chi_{2}}{1.7\times10^{-5}}=\dfrac{300}{150}$

$\chi_{2}=\dfrac{300\times1.7\times10^{-5}}{150}$

$\chi_{2}=3.4\times10^{-5}$

We need to calculate the magnetization

Using formula of magnetization

$I=\chi_{2}\times H$

Where, H = magnetic intensity

Formula of magnetic intensity

$H=\dfrac{B}{\mu_{0}}$

Where, B = magnetic field

Put the value of H into the formula of magnetization

$I=\chi_{2}\times\dfrac{B}{\mu_{0}}$

$I=3.4\times10^{-5}\times\dfrac{1.5}{4\pi\times10^{-7}}$

$I=40.6\ A/m$

Hence, The magnetization of a small sample of aluminum is 40.6 A/m