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(B) The total dipole moment of the sample at saturation is $M_{	ext{sat}} = N \mu = (2 	imes 10^{24}) 	imes (1.5 	imes 10^{-23} 	ext{ J T}^{-1})

Vedclass · Accepted Answer

$13.5$

Vedclass · Answer

(B) The total dipole moment of the sample at saturation is $M_{	ext{sat}} = N \mu = (2 	imes 10^{24}) 	imes (1.5 	imes 10^{-23} 	ext{ J T}^{-1}) = 30 	ext{ J T}^{-1}$.According to Curie's law for paramagnetism, the magnetization $M$ is proportional to $B/T$, i.e., $M \propto B/T$.In the first case, the degree of saturation is $20 \%$, so the magnetic moment is $M_1 = 0.20 	imes M_{	ext{sat}} = 0.20 	imes 30 = 6 	ext{ J T}^{-1}$.The ratio of magnetic moments is given by $\frac{M_2}{M_1} = \frac{B_2 / T_2}{B_1 / T_1}$.Substituting the values: $\frac{M_2}{6} = \frac{0.9 / 2.8}{0.6 / 4.2} = \frac{0.9}{2.8} 	imes \frac{4.2}{0.6} = \frac{0.9}{0.6} 	imes \frac{4.2}{2.8} = 1.5 	imes 1.5 = 2.25$.Therefore, $M_2 = 6 	imes 2.25 = 13.5 	ext{ J T}^{-1}$.

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