Graph $$\left[ A \right]$$ is for material used for making permanent magnets (high coercivity)
Graph $$\left[ B \right]$$ is for making electromagnets and transformers.
2.
If the angular momentum of an electron is $$\vec J$$ then the magnitude of the magnetic moment will be
We know that,
$$\eqalign{
& \frac{M}{J} = \frac{q}{{2m}} \cr
& \therefore M = J \times \frac{e}{{2m}}. \cr} $$
3.
A bar magnet has coercivity $$4 \times {10^3}A{m^{ - 1}}.$$ It is desired to demagnetise it by inserting it inside a solenoid $$12\,cm$$ long and having $$60$$ turns. The current that should be sent through the solenoid is
4.
A permanent magnet in the shape of a thin cylinder of length $$10\,cm$$ has magnetisation $$\left( M \right) = {10^6}\,A\,{m^{ - 1}}.$$ Its magnetization current $${I_M}$$ is
5.
A magnet makes $$40$$ oscillations per minute at a place having magnetic field intensity of $$0.1 \times {10^{ - 5}}T.$$ At another place, it takes $$2.5\,\sec$$ to complete one vibration. The value of earth’s horizontal field at that place is
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.
7.
The net magnetic moment of two identical magnets each of magnetic moment $${M_0},$$ inclined at $${60^ \circ }$$ with each other is
8.
Two identical short bar magnets, each having magnetic moment of $$10\,A{m^2},$$ are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane, If the distance between their centres is $$0.2\,m,$$ the resultant magnetic induction at a point midway between them is $$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}H{m^{ - 1}}} \right)$$
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.
10.
A magnet of magnetic moment $$50\hat i\,A - {m^2}$$ is placed along the $$x$$-axis in a magnetic field $$\vec B = \left( {0.5\hat i + 3.0\hat j} \right)T.$$ The torque acting on the magnet is