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Chapter 20: Problem 40

A paramagnetic sample shows a net magnetisation of \(8 \mathrm{Am}^{-1}\) when placed in an external magnetie field of \(0.6 \mathrm{~T}\) at a temperature of \(4 \mathrm{~K}\). When the same sample is placed in an external magnetic field of \(0.2 \mathrm{~T}\) at a temperature of \(16 \mathrm{~K}\), the magnetisation will be [NCFRT Exemplar] (a) \(\frac{32}{3} \mathrm{Am}^{-1}\) (b) \(\frac{2}{3} \mathrm{Am}^{-1}\) (c) \(6 \mathrm{Am}^{-1}\) (d) \(2.4 \mathrm{Am}^{-1}\)

Short Answer

Expert verified
The magnetisation will be \(\frac{2}{3} \mathrm{Am}^{-1}\). Correct answer: (b).

Step by step solution

01

Understanding Curie's Law

According to Curie's law, for a paramagnetic material, the magnetisation (\(M\)) is directly proportional to the magnetic field (\(B\)) and inversely proportional to the temperature (\(T\)). The relation is given by \[ M = C \frac{B}{T} \] where \(C\) is the Curie constant.
02

Write the Equation for Initial Condition

For the initial condition, the sample has a magnetisation \(M_1 = 8 \mathrm{Am}^{-1}\) in a magnetic field \(B_1 = 0.6 \mathrm{~T}\) at a temperature \(T_1 = 4 \mathrm{~K}\). Using Curie's law, we have \[ M_1 = C \frac{B_1}{T_1} \] which implies \[ 8 = C \frac{0.6}{4} \].
03

Solve for Curie Constant C

Rearrange the equation to solve for \(C\):\[ C = \frac{8 \times 4}{0.6} = \frac{32}{0.6} \approx 53.33 \mathrm{Am}^{-1} \cdot \mathrm{K}/\mathrm{T} \].
04

Write Equation for New Condition

For the changed condition, we have the magnetisation \(M_2\) in a magnetic field \(B_2 = 0.2 \mathrm{~T}\) at a temperature \(T_2 = 16 \mathrm{~K}\). Using Curie's law, \[ M_2 = C \frac{B_2}{T_2} \] which implies \[ M_2 = 53.33 \frac{0.2}{16} \].
05

Calculate New Magnetisation M2

Simplify and calculate \(M_2\) using the values of \(C\), \(B_2\), and \(T_2\):\[ M_2 = 53.33 \times \frac{0.2}{16} = \frac{53.33 \times 0.2}{16} \approx 0.6666 \mathrm{Am}^{-1} \].
06

Round and Match Answer

The calculated magnetisation \(M_2\) is approximately \(0.6666 \mathrm{Am}^{-1}\), which can be written as \(\frac{2}{3} \mathrm{Am}^{-1}\). Hence, the correct answer is option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paramagnetic Materials
Paramagnetic materials are a fascinating category of materials that display magnetic behavior under certain conditions. These materials have unpaired electrons that contribute to their magnetic moments. However, unlike ferromagnetic materials, paramagnetic materials do not retain any magnetization in the absence of an external magnetic field.
This key property means that their magnetization depends heavily on the external magnetic field applied to them.
Because of their unique electron arrangement:
  • They align with external magnetic fields.
  • They exhibit a positive but weak attraction to magnetic fields.
  • Their magnetization strengthens with increasing magnetic fields and decreases with higher temperatures.
Understanding the inherent properties of paramagnetic materials is crucial when applying concepts like Curie's Law, as it helps in predicting how these materials will behave in different conditions of temperature and magnetic field strength.
Magnetisation
Magnetisation is the measure of the magnetic moment per unit volume of a material. It indicates how strongly a material reacts to an external magnetic field.
In paramagnetic materials, magnetisation is not intrinsic, which means it only appears when an external magnetic field is applied.
Using Curie's Law, the magnetisation, denoted as \(M\), can be calculated as:\[ M = C \frac{B}{T} \]where \(C\) is the Curie constant, \(B\) is the external magnetic field, and \(T\) is the temperature.
This formula shows that:
  • Magnetisation increases with a stronger magnetic field *\(B\)*.
  • Lower temperatures support higher magnetisation.
  • The Curie constant *\(C\)* is material-specific and affects the level of magnetisation proportionally.
Thus, by understanding these relationships, one can predict how a paramagnetic sample will behave under different environmental conditions.
Magnetic Field
A magnetic field is an invisible force field created by moving electric charges and intrinsic magnetic moments of elementary particles. It exerts forces on other moving charges and magnetic dipoles, like those found in paramagnetic materials.
Magnetic fields are represented by field lines extending from the north to the south pole of a magnet.
In terms of paramagnetic materials, the influence of an external magnetic field is critical for inducing magnetisation.
This impact reflects:
  • The alignment of atomic magnetic moments along the field direction.
  • Strengthening of magnetisation with increased field strength.
  • Dependence on the thermal environment, as hotter atoms misalign more easily than cooler ones.
Through this interaction with an external magnetic field, one can directly observe how the inherent properties of paramagnetic materials are influenced, helping to derive useful predictions using Curie's Law.

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Most popular questions from this chapter

In a certain place, the horizontal component of magnetic field is \(\frac{1}{\sqrt{3}}\) times the vertical component. The angle of dip at this place is (a) zero (b) \(\pi / 3\) (c) \(\pi / 2\) (d) \(\pi / 6\)

Match the following column I with column II Column I Column II I. Magnetic moment A. \(\left[\mathrm{ML}^{-} \mathrm{T}^{-2} \mathrm{~A}^{-1}\right]\) II. Permeability B. Vector III. Intensity of magnetisation C. \(\mathrm{Nm}^{\mathrm{I}} / \mathrm{Wb}\) IV. Magnetic induction D. Scalar Code (a) \(1-\mathrm{A}, \mathrm{II}-\mathrm{B}, \mathrm{III}-\mathrm{C}, \mathrm{N}-\mathrm{D}\) (b) \(1-C, I I-D, I \|-B, I V-A\) (c) \(1-\mathrm{D}, \mathrm{II}-\mathrm{C}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}\) (d) \(\mathrm{L}-\mathrm{B}, \mathrm{Il}-\mathrm{A}, \mathrm{m}-\mathrm{B}, \mathrm{N}-\mathrm{D}\)

A magnet. \(20 \mathrm{~cm}\) long with its poles concentrated at its ends is placed vertically with its north pole on the table. At a point due \(20 \mathrm{~cm}\) south (magnetic) of the pole, a neutral point is obtained. If \(H=0.3 \mathrm{G}\), then the pole strength of the magnet is approximately (a) \(185 \mathrm{ab}-\mathrm{amp}-\mathrm{cm}\) (b) \(185 \mathrm{amp}-\mathrm{m}\) (c) \(18.5 \mathrm{ab}-\mathrm{amp-cm}\) (d) \(18.5 \mathrm{amp-m}\)

A horizontal circular loop carries a current that looks anti-clockwise when viewed from above. It is replaced by an equivalent magnetic dipole \(N \cdot S\). Which of the following is true? (a) The line \(N\)-S should be along a diameter of the loop (b) The line \(N\) - should be perpendicular to the plane of the loop (c) South pole should be below the loop (d) North pole should be below the loop

A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is \(\theta\) and the period of oscillation of the needle in the magnetometer is \(T\). When the magnet is removed, the period of oscillation is \(T_{0}\). Find the relation between \(T\) and \(T_{0}\) is (a) \(T^{2}=T_{0}^{2} \cos \theta\) (b) \(T=T_{0} \cos \theta\) (c) \(T=\frac{T_{0}}{\cos \theta}\) (d) \(T^{2}=\frac{T_{0}^{2}}{\cos \theta}\)
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