Question

A paramagnetic sample shows a net magnetization of 6 Am’1 when placed in an external magnetic field of 0.4 T at a temperature of 4 K. When the same sample is placed in an external magnetic field of 0.3 T at a temperature of 24 K, the magnetization will be

• A. 0.75 Am-1
• B. 1.0 Am-1
• C. 1.25 Am-1
• D. 1.5 Am-1

Answer 1

Answer: (A)

0.75 Am-1

Answer 2

For a paramagnetic material, the magnetization (M) is directly proportional to the applied magnetic field (B) and inversely proportional to the absolute temperature (T). This relationship is described by Curie's Law:

$M = C \frac{B}{T}$

where C is Curie's constant.

Given the initial conditions (Case 1):

$M_1 = 6 \text{ Am}^{-1}$

$B_1 = 0.4 \text{ T}$

$T_1 = 4 \text{ K}$

Given the final conditions (Case 2):

$B_2 = 0.3 \text{ T}$

$T_2 = 24 \text{ K}$

We need to find $M_2$.

From Curie's Law, we can write for both cases:

$M_1 = C \frac{B_1}{T_1} \quad \text{(1)}$

$M_2 = C \frac{B_2}{T_2} \quad \text{(2)}$

Divide equation (2) by equation (1):

$\frac{M_2}{M_1} = \frac{C \frac{B_2}{T_2}}{C \frac{B_1}{T_1}}$

$\frac{M_2}{M_1} = \frac{B_2}{T_2} \times \frac{T_1}{B_1}$

Now, substitute the given values:

$M_2 = M_1 \times \frac{B_2}{B_1} \times \frac{T_1}{T_2}$

$M_2 = 6 \text{ Am}^{-1} \times \frac{0.3 \text{ T}}{0.4 \text{ T}} \times \frac{4 \text{ K}}{24 \text{ K}}$

$M_2 = 6 \times \frac{3}{4} \times \frac{1}{6}$

$M_2 = \frac{6 \times 3 \times 1}{4 \times 6}$

$M_2 = \frac{18}{24}$

$M_2 = \frac{3}{4}$

$M_2 = 0.75 \text{ Am}^{-1}$

Therefore, the magnetization will be $0.75 \text{ Am}^{-1}$.