# Concept Of Physics Magnetic Properties of Matter

#### H C Verma

1.   The magnetic intensity H at the centre of a long solenoid carrying a current of $2\cdot0\ A$, is found to be $1500\ A/m$. Find the number of turns per centimetre of the solenoid.

$B = \mu_{0}ni$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ $H = \dfrac{B}{\mu_{0}}$ $\\$ $\Rightarrow H = ni$ $\\$ $\Rightarrow 1500\ A/m = n \times 2$ $\\$ $\Rightarrow n = 750\ Turns/meter$ $\\$ $\Rightarrow n = 7.5\ turns/cm$

2.   A rod is inserted as the core in the current-carrying solenoid of the previous problem, (a) What is the magnetic intensity $H$ at the centre ? (b) If the magnetization / of the core 'is found to be $0\cdot12\ A/m$, find the susceptibility of the material of the rod. (c) Is the material paramagnetic, diamagnetic or ferromagnetic ?

a) $H = 1500\ A/m$ $\\$ As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, the end effects may be neglected. There is no effect of the rod on the magnetic intensity at the centre. $\\$ (b) $I = 0.12\ A/m$ $\\$ We know $\vec{I} = X\vec{H} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X =$ Susceptibility $\\$ $\Rightarrow X = \dfrac{I}{H} = \dfrac{0.12}{1500} = 0.00008 = 8 \times 10^{-5}$ $\\$ (c) The material is paramagnetic

3.   The magnetic field inside a long solenoid having $50\ turns/cm$ is increased from $2\cdot5\ \times 10^{-3}\ T$ to $2\cdot5\ T$ when an iron core of cross-sectional area $4\ cm$ is inserted into it. Find (a) the current in the solenoid, (b) the magnetization $I$ of the core and (c) the pole strength developed in the core.

$B_1 = 2.5 \times 10^{–3}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B_2 = 2.5$ $\\$ $A = 4 \times 10^{–4}\ m^2, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 50\ turns/cm = 5000\ turns/m$ $\\$ a) $B = \mu_{0}ni$, $\\$ $\Rightarrow 2.5 \times 10^{-3} = 4\pi \times 10^{-7} \times 5000 \times i$ $\\$ $\Rightarrow i = \dfrac{2.5 \times 10^{-3}}{4\pi \times 10^{-7} \times 5000} = 0.398 A \approx 0.4 A$

b) $I = \dfrac{B_2}{\mu_0} - H = \dfrac{2.5}{4\pi \times 10^{-7}} - (B_2 - B_1) = \dfrac{2.5}{4\pi \times 10^{-7}} - 2.497 = 1.99 \times 10^{6} \approx 2 \times 10^{6}$ $\\$ c) $I = \dfrac{M}{V} \Rightarrow I = \dfrac{ml}{Al} = \dfrac{m}{A}$ $\\$ $\Rightarrow m = IA = 2 × 10^6 \times 4 \times 10^{–4} = 800\ A-m$

4.   A bar magnet of length $1\ cm$ and cross-sectional area $1\cdot0\ cm$ produces a magnetic field of $1\cdot5 \times 10^{-4}\ T$ at a point in end-on position at a distance $15\ cm$ away from the centre, (a) Find the magnetic moment M of the magnet, (b) Find the magnetization $I$ of the magnet, (c) Find the magnetic field $B$ at the centre of the magnet.

a) Given $d = 15\ cm = 0.15\ cm$

$\ell = 1\ cm = 0.01\ m$

$A = 1.0\ cm^2 = 1 \times 10^{-4}\ m^2$ $\\$ $B = 1.5 \times 10^{-4}\ T$ $\\$ $m = ?$

We know $\vec{B} = \dfrac{\mu_{0}}{4\pi} \times \dfrac{2Md}{(d^2 - \ell^2)^2}$ $\\$ $\Rightarrow 1.5 \times 10{-4} = \dfrac{10^{-7} \times 2 \times M \times 0.15}{(0.0225 - 0.0001)^2} = \dfrac{3 \times 10^{-8}M}{5.01 \times 10^{-4}}$ $\\$ $\Rightarrow M = \dfrac{1.5 \times 10^{-4} \times 5.01 \times 10^{-4}}{3 \times 10^{-8}} = 2.5 A$ $\\$ b) Magnetisation $I = \dfrac{M}{V} = \dfrac{2.5}{10^{-4} \times 10^{-2}} = 2.5 \times 10^{6}\ A/m$ $\\$ c) $H = \dfrac{m}{4\pi d^2} = \dfrac{M}{4\pi Id^2} = \dfrac{2.5}{4 \times 3.14 \times 0.01 \times (0.15)^2}$ $\\$ net $H = H_{N} + H = 2 \times 884.6 = 8.846 \times 10^2$ $\\$ $\vec{B} = \mu_{0}(-H + I) = 4\pi \times 10^{-7} (2.5 \times 10^{6} - 2 \times 884.6) \approx 3,14\ T$

5.   The susceptibility of annealed iron at saturation is $5500$. Find the permeability of annealed iron at saturation.

Permeability $(\mu) = \mu_{0}(1 + x)$ $\\$ Given susceptibility $= 5500$ $\\$ $\mu = 4 \times 10^{-7} (1 + 5500)$ $\\$ $= 4 \times 3.14 \times 10^{-7} \times 5501\ 6909.56 \times 10^{-7} \approx 6.9 \times 10^{-3}$

6.   The magnetic field $B$ and the magnetic intensity $H$ in a material are found to be $1\cdot6\ T$ and $1000\ A/m$ respectively. Calculate the relative permeability $\mu$, and the susceptibility $x$ of the material.

$B = 1.6\ T,\ H = 1000\ A/m$ $\\$ $\mu =$ Permeability of material $\\$ $\mu = \dfrac{B}{H} =\dfrac{1.6}{1000} = 1.6 \times 10^{-3}$ $\\$ $\mu{r} = \dfrac{\mu}{\mu_{0}} = \dfrac{1.6 \times 10^{-3}}{4\pi \times 10^{-7}} = 0.127 \times 10^{4} \approx 1.3 \times 10^{3}$ $\\$ $\mu = \mu_{0}(1 + x)$ $\\$ $\Rightarrow x =\dfrac{\mu}{\mu_{0}} - 1$ $\\$ $= \mu_{r} - 1 = 1.3 \times 10^{3} – 1 = 1300 – 1 = 1299 \approx 1.3 \times 10^{3}$

7.   The susceptibility of magnesium at $300\ K$ is $1\cdot2 \times 10^{-6}$ . At what temperature will the susceptibility increase to $1\cdot8 \times 10^{-5}$ ?

$x = \dfrac{C}{T} = \Rightarrow \dfrac{x_1}{x_2} = \dfrac{T_2}{T_{1}}$ $\\$ $\Rightarrow \dfrac{1.2 \times 10^{-5}}{1.8 \times 10^{-5}} = \dfrac{T_2}{200}$ $\\$ $\Rightarrow T_2 = \dfrac{12}{18} \times 300 = 200\ K$

8.   Assume that each iron atom has a permanent magnetic moment equal to $2$ Bohr magnetons $(1$ Bohr magneton equals $9\cdot27 \times 10^{-24}\ A-m^2)$. The density of atoms in iron is $8\cdot52 \times 10^{28}\ atoms/m^3$. (a) Find the maximum magnetization $I$ in a long cylinder of iron, (b) Find the maximum magnetic field $B$ on the axis inside the cylinder.

$f = 8.52 × 10^{28} atoms/m^3$ $\\$ For maximum $‘I’$, Let us consider the no. of atoms present in $1\ m^3$ of volume. $\\$ Given: $m$ per atom $= 2 \times 9.27 \times 10^{–24} A–m^2.$ $\\$ $I = \dfrac{net\ m}{V} = 2 \times 9.27 \times 10^{-24} \times 8.52 \times 10^{28} \approx 1.58 \times 10^{6}\ A/m$ $\\$ $B = \mu_{0}(H + 1) = \mu_{0}I \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ [$\therefore H = 0$ in this case] $\\$ $= 4\pi \times 10^{-7} \times 1.58 \times 10^{6} = 1.98 \times 10^{-1} \approx 2.0\ T$

9.   The coercive force for a certain permanent magnet is $4\cdot0\ \times 10\ A/m$. This magnet is placed inside a long solenoid of $40\ turns/cm$ and a current is passed in the solenoid to demagnetise it completely. Find the current.

$B =\mu_{0}ni, \ \ \ \ \ \ \ \ \ \ H = \dfrac{B}{\mu_{0}}$ $\\$ Given $n = 40$ turn/cm $= 4000$ turns/m $\\$ $\Rightarrow H = ni$ $\\$ $H = 4 \times 10^{4}\ A/m$ $\\$ $\Rightarrow i = \dfrac{H}{n} = \dfrac{4 \times 10^{4}}{4000} = 10\ A$