Test Index

NCERT Class XII Chapter
Magnetism and Matter
Questions With Solutions

© examsnet.com
Question : 23 of 25
Marks: +1, -0
A sample of paramagnetic salt contains 2.0 × 102410^{24} atomic dipoles each of dipole moment 1.5 × 10−2310^{-23} J T−1T^{-1}. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
Solution:  
Dipole moment at complete saturation
M = 2 × 102410^{24} × 1.5 × 10−2310^{-23} = 30 J T−1T^{-1}
At 0.84 T and 4.2 K, 15% of sample is magnetised
M = C B0T0\frac{B_0}{T_0}
0.15 × 30 = C 0.844.2\frac{0.84}{4.2} ... (i)
At 0.98 T and 2.8 K
M = C 0.982.8\frac{0.98}{2.8}
Dividing (i) by (ii), M = 7.875 J T−1T^{-1}
© examsnet.com
Go to Question:
Prev Question
Next Question