When temperature of a ferromagnetic material is increased by 20%, what is the percentage change in its magnetic susceptibility?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 16.7%
The magnetic susceptibility \( \chi \) measures how strongly a material is magnetised by an applied field (\( M = \chi H \)). The magnetic order in the material is disturbed by thermal agitation, so the susceptibility falls as the temperature rises. The question expects the simple inverse dependence of Curie's law, \( \chi \propto 1/T \) (for a ferromagnet above its Curie temperature the fuller form is \( \chi \propto 1/(T - T_C) \), but no \( T_C \) is given, so the inverse law is used):
\( \displaystyle \frac{\chi_2}{\chi_1} = \frac{T_1}{T_2} \)
The temperature rises by 20%, so \( T_2 = 1.2\,T_1 \):
\( \displaystyle \frac{\chi_2}{\chi_1} = \frac{1}{1.2} = 0.833 \)
\( \displaystyle \text{Percentage change} = \left(\frac{\chi_2}{\chi_1} - 1\right) \times 100\% = (0.833 - 1) \times 100\% \approx -16.7\% \)
The susceptibility decreases by about 16.7%.