Magnetic Materials MCQ Questions & Answers in Electrostatics and Magnetism | Physics

Graph $$\left[ A \right]$$  is for material used for making permanent magnets (high coercivity)
Graph $$\left[ B \right]$$  is for making electromagnets and transformers.

We know that,
$$\eqalign{ & \frac{M}{J} = \frac{q}{{2m}} \cr & \therefore M = J \times \frac{e}{{2m}}. \cr} $$

$$\eqalign{ & n = \frac{N}{l} = \frac{{60}}{{0.12}} = 500\,{\text{turn}}\,metr{e^{ - 1}} \cr & \Rightarrow i = \frac{H}{n} = \frac{{4 \times {{10}^3}}}{{500}} = 8A \cr} $$

$${I_M} = ML = {10^6} \times 0.1 = {10^5}A$$

$${T_1} = \frac{{60}}{{40}} = 1.5\,s\,{\text{and}}\,{T_2} = 2.5\,s$$
We know that, $$\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{B_2}}}{{{B_1}}}} $$
$$\eqalign{ & {\text{or}}\,{B_2} = \frac{{T_1^2}}{{T_2^2}} \times {B_1} = {\left( {\frac{{1.5}}{{2.5}}} \right)^2} \times \left( {0.1 \times {{10}^{ - 5}}} \right) \cr & = 0.36 \times {10^{ - 6}}T \cr} $$

Nickel exhibits ferromagnetism because of a quantum physical effect called exchange coupling in which the electron spins of one atom interact with those of neighbouring atoms. The result is alignment of the magnetic dipole moments of the atoms, in spite of the randomising tendency of atomic collisions. This persistent alignment is what gives ferromagnetic materials their permanent magnetism.
If the temperature of a ferromagnetic material is raised above a certain critical value, called the Curie temperature, the exchange coupling ceases to be effective.
Most such materials become simply paramagnetic, i.e. the dipoles still tend to align with an external field but much more weakly and thermal agitation can now more easily disrupt the alignment.

$$\eqalign{ & {M_{{\text{net}}}} = \sqrt {M_0^2 + M_0^2 + 2M_0^2\cos {{60}^ \circ }} \cr & = \sqrt {3M_0^2} = \sqrt 3 {M_0} \cr} $$

From figure $${B_{{\text{net}}}} = \sqrt {B_a^2 + B_e^2} $$
$$ = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}}} \right)}^2} + {{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}}} \right)}^2}} $$
$$ = \sqrt 5 .\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}} = \sqrt 5 \times {10^{ - 7}} \times \frac{{10}}{{{{\left( {0.1} \right)}^3}}} = \sqrt 5 \times {10^{ - 3}}\,tesla$$

Earth's magnetic field at poles is vertical (perpendicular to the earth's surface) and horizontal (parallel to the earth’s surface) at equator.
Cosmic rays are positively charged particles and its velocity is parallel to the earths magnetic field. So, no magnetic force acts on cosmic ray particles coming at poles, i.e. force
$$\left( F \right) = qvB\sin \theta $$
At poles angle between $$v$$ and $$B$$ is $$\theta = {0^ \circ },$$
So, $$F = 0$$
At equator $$\theta = {90^ \circ },$$
So, $${\text{Force}} = qvB\sin {90^ \circ } = qvB$$
This force is maximum and deflects the particles sideways.
Hence, only high energy particles can reach the equator.

$$\vec \tau = \vec M \times \vec B = 50\,\hat i \times \left( {0.5\hat i + 3\hat j} \right) = 150\,\hat k\,N{\text{ - }}m$$