Solutions for Exercises of Magnetic Materials
Q.3 When a plate of magnetic material of size 10 cm × 5 cm × 0.2 cm (length, breadth and thickness respectively) is located in magnetizing field of 0.5 × 104 Am-1 then a magnetic moment of 5 Am2 is induced in it. Find out magnetic induction in the rod.
V = 10 cm × 0.5cm × 0.2cm = 1cm3 = 10-6m3
H = 0.5 × 104Am-1, Mnet = 5Am2, B = ?
Magnetization of the material is given as,
$latex M=\frac{M_{net}}V$
$latex \frac5{10^{-6}}=5\times10^6A/m$
Relation between magnetization and magnetizing field is given as,
$latex \Rightarrow X=\frac MH$ Here, X is a susceptibility
$latex \therefore X=\frac MH=\frac{5\times10^6}{0.5\times10^4}$
$latex =10^3=1000$
$latex \therefore\;\mu_r\;=\;1\;+\;\chi\;=\;1001\;$
$latex \therefore\;B\;=\;\mu H\;=\;\mu_0\mu_rH$
$latex \therefore\;B\;=\;4\pi\;\times\;10^{-7}\;\times\;1001\;\times\;0.5\;\times\;10^4\;$
$latex \therefore\;B\;=\;6.34\;T$
4) A rod of magnetic material of cross section 0.25 cm2 is located in 4000 Am-1 magnetizing field. Magnetic flux passing through the rod is 25 x 10-6 Wb. Find out
(a) relative permeability
(b) magnetic susceptibility and
(c) magnetisation of the rod
A = 0.25cm2 = 0.25 × 10-4m2, H = 4000 Am-1,
ϕ = 25 × 10-6Wb, μr = ?,χ = ?,M =?
Magnetic induction or flux density inside the rod is,
$latex \therefore\;B\;=\frac\phi A$
$latex \therefore\;B=\frac{25\times10^{-6}}{0.25\times10^{-4}}$
$latex \therefore B=1Wb/m^2$
$latex B\;=\;\mu H\;\;\therefore\;\mu\;=\;\frac BH$
$latex =\frac1{4000}=2.5\times10^{-4}$
$latex \mu=\mu_0\mu_r\;\;\;\;\therefore\;\mu_r=\frac\mu{\mu_0}\;$
$latex =\frac{2\pi\times10^{-4}}{4\pi\times10^{-7}}=199$
But, μr = 1 + χ ∴ χ = μr − 1 = 198
$latex \chi=\frac HM\;\;\;\;\;\;\therefore M=XH$
$latex =\;198\;\times\;4000$
$latex =7.92\;\times\;10^5\;A/m$
5) The work done for rotating a magnet with magnetic dipole moment m, through 90° from its magnetic meridian is n times the work done to rotate it through 60°. Find the value of n.
Let W1 is work for ????1 = 90° and W2 for ????2 = 60°
Magnetic meridian is the equilibrium position of magnet
i.e. ???? = 0°
Work required to rotate magnet (from equilibrium) in magnetic field is given by,
$latex \Rightarrow\;W\;=\;mB(1\;-\;\cos\theta)$
$latex \therefore\;W_1\;=\;mB(1\;-\;\cos90)$
$latex i.e\;W_1\;=\;mB$
$latex \therefore\;W_2\;=\;mB(1\;-\;\cos60)$
$latex i.e\;W_2\;=\;mB\left(1-\frac12\right)=\frac{mB}2$
Given that, W1 = nW2
$latex \therefore\;n\;=\;\frac{W_1}{W_2}=\;\frac{mB}{\displaystyle\frac{mB}2}\;=\;2$
6) An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 x 10-11 m, with a speed of 2 x 106 ms-1. Find the resultant orbital magnetic moment and angular momentum of electron. (charge on electron e = 1.6 x 10-19 C, mass of electron m = 9.1 x 10-31 kg.)
r = 5.3 × 10-1 m, v = 2 × 106ms-1, more = ?, L = ?
$latex m_{orb}=\frac{evr}2$
$latex =\frac{1.6\times10^{-1}\times2\times10^6\times5.3\times10^{-11}}2$
$latex \therefore\;m_{orb}\;=\;8.48\;\times\;10^{-24}Am^2\;$
L = mvr = 9.1 × 10-31 × 2 × 106 × 5.3 × 10-1
∴ L = 9.65 × 10-35kgm2/s
7) A paramagnetic gas has 2.0 x 1026 atoms/m3 with atomic magnetic dipole moment of 1.5 x 10-23 A-m2 each. The gas is at 27° C. (a) Find the maximum magnetization intensity of this sample. (b) If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why? (Hint: Find the ratio of Thermal energy of atom of a gas ( 3/2 kBT) and maximum potential energy of the atom (mB) and draw your conclusion)
N = 2 × 1026, m= 1.5 × 10-2 Am2, T = 27°C = 300K
M = ?, Is saturated if kept in B = 2 T
As moment of each atom is given, maximum possible net moment (M) can be given as,
M = N × m
M= 2 × 1026 × 1.5 × 10-23
∴ M = 3 × 103Am-1
Thermal energy of atom of a gas is given as,
$latex E_{thermal}=\frac32k_BT$
$latex =\frac32\times1.38\times10^{-23}\times300$
$latex \therefore\;E_{thermal}\;=\;621\;\times\;10^{-23}J\;$
8) A magnetic needle placed in uniform magnetic field has magnetic moment of 2 x 10-2 Am2, and moment of inertia of 7.2 x 10-7 kg m2. It performs 10 oscillations in 6s. What is the magnitude of the magnetic field?
m = 2 × 10-2Am2, I = 7.2 × 10-7kgm2
$latex T\;=\frac6{10}=0.6\;s$ , B=?
We know that for vibration magnetometer,
$latex T\;=\;2\pi\sqrt{\frac I{mB}}$
squaring both the sides. We get,
$latex T^2=4\pi^2\left(\frac I{mB}\right)\;\;\;\;\;\;\therefore B=\frac{4\pi^2I}{mT^2}\;\;\;$
$latex \therefore\;B\;=\;\frac{4\times3.14^2\times7.2\times10^{-7}}{2\times10^{-2}\times0.6\times0.6}$
$latex \therefore\;B\;=\;39.48\;\times\;10^{-4}\;T$
9) The susceptibility of a paramagnetic material is X at 27°C. At what temperature its susceptibility be X/3 ?
χ1 = χ, χ2 = χ/3, T1 = 27°C = 300K, T2=?
For paramagnetic material,
We have,
$latex \chi\;\alpha\;\frac1T\left(in\;K\right)\;\;\;\;\;\;\therefore\frac{\chi_1}{\chi_2}=\frac{T_2}{T_1}$
$latex \therefore T_2=\frac{\chi_1}{\chi_2}\times T_1$
$latex \therefore T_2=\frac{\chi\;}{\displaystyle\frac{\chi\;}3}\times300$
= 900K = 627℃