A paramagnetic sample shows a net magnetizati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Curie's Law,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$,i.e.,$\chi = \frac{C}{T}$.Th

Vedclass · Accepted Answer

$\frac{2}{3} \ Am^{-1}$

Vedclass · Answer

(B) According to Curie's Law,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$,i.e.,$\chi = \frac{C}{T}$.The magnetization $I$ is given by $I = \chi H$,where $H$ is the magnetic field intensity. Since $B = \mu_0 H$,we have $H = \frac{B}{\mu_0}$.Thus,$I = \chi \frac{B}{\mu_0} = \frac{C}{T} \cdot \frac{B}{\mu_0} = K \frac{B}{T}$,where $K$ is a constant.Given:$I_1 = 8 \ Am^{-1}$,$B_1 = 0.6 \ T$,$T_1 = 4 \ K$$I_2 = ?$,$B_2 = 0.2 \ T$,$T_2 = 16 \ K$Using the ratio $\frac{I_2}{I_1} = \frac{B_2}{B_1} \cdot \frac{T_1}{T_2}$:$\frac{I_2}{8} = \frac{0.2}{0.6} \cdot \frac{4}{16}$$\frac{I_2}{8} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$$I_2 = 8 \cdot \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \ Am^{-1}$.

$A$ paramagnetic sample shows a net magnetization of $8 \ Am^{-1}$ when placed in an external magnetic field of $0.6 \ T$ at a temperature of $4 \ K$. When the same sample is placed in an external magnetic field of $0.2 \ T$ at a temperature of $16 \ K$,the magnetization will be

Similar Questions

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

Which of the following amines undergoes acylation reaction?

In angiosperms,the functional megaspore develops into which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module