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A sample of paramagnetic salt contains 2.0xx 10^(24) atomic dipoles each of dipole moment 1.5 xx 10^(-23) J T^(-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) |
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Answer» Solution :INDUCED MAGNETIC DIPOLE moment obtained in a given paramagnetic sample of volume V at temperature `T_1` is, `(m_b)_(1) = 15%` of `(m_b)_("max")` `= 15%` of `Nm` `= (2 xx 10^(24) ) (1.5 xx 10^(-23) ) ((15)/( 100))` `= 4.5 Am^(2) ""...(1)` `M= (m_b)/( V) = ("total induced dipole moment")/("total volume")` `therefore (M_1)/( M_2) = ((m_b)_(1) ) /( V)xx (V)/( (m_b)_(2) )= ((m_b)_(1) )/( (mb)_(2) ) ""...(2)` Magnetisation obtained is given by Curie.s law, `M= C(B)/( T)` (Where `C=` Curie.s constant) `rArr M prop (B)/(T)` `therefore (M_1)/( M_2) = (B_1)/( B_2) xx (T_2)/( T_1) ` `therefore ((m_b)_(1) )/( (m_b)_(2) ) = (B_1 )/( B_2) xx (T_2)/( T_1)` `therefore (m_b)_(2) = (m_b)_(1) xx (B_2)/(B_1) xx (T_1)/( T_2)` `therefore (m_b)_(2) = 4.5 xx (0.98)/( 0.84) xx(4.2)/(2.8) = 7.875 "Am"^*(-1)` (or `JT^(-1)` ) |
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