Solution: (i) Magnetic field lines form continuous closed loops. (ii) The tangent to the field line at a given point represents the direction of the net magnetic field at that point. (iii) Two magnetic field lines never intersect each other.
Solution: Magnetic field lines form continuous loops because isolated magnetic monopoles do not exist. Every north pole is associated with a south pole. Electric field lines start at positive charges and end at negative charges because isolated electric charges exist.
Solution: For a short bar magnet ($r \gg l$):
Axial line: $B_{axial} = \frac{\mu_0}{4\pi} \frac{2m}{r^3}$
Equatorial line: $B_{equatorial} = \frac{\mu_0}{4\pi} \frac{m}{r^3}$
Solution: A solenoid with current $I$ acts as a series of magnetic dipoles. Its magnetic field pattern exterior to the solenoid is identical to that of a bar magnet.
Similarity: Both experience torque in an external magnetic field and align N-S when suspended freely.
Difference: The magnetic field strength of a solenoid can be varied by changing the current, whereas a bar magnet's strength is fixed.
Solution: Torque $\vec{\tau} = \vec{m} \times \vec{B}$ (Magnitude: $\tau = mB\sin\theta$). The torque is maximum when the dipole is held perpendicular to the magnetic field ($\theta = 90^\circ$), and $\tau_{max} = mB$.
Solution: Given: $\theta = 30^\circ$, $B = 0.25\text{ T}$, $\tau = 4.5 \times 10^{-2}\text{ J}$.
$\tau = mB\sin\theta \Rightarrow m = \frac{\tau}{B\sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} = 0.36\text{ J/T}$ (or $\text{A m}^2$).
Solution: Stable equilibrium means initial angle $\theta_1 = 0^\circ$. Final angle $\theta_2 = 60^\circ$.
Work $W = mB(\cos\theta_1 - \cos\theta_2) = (3.0)(0.1)(\cos 0^\circ - \cos 60^\circ) = 0.3(1 - 0.5) = 0.3 \times 0.5 = 0.15\text{ Joules}$.
Solution: Potential Energy $U = -mB\cos\theta$.
(i) Minimum when $\theta = 0^\circ$ (dipole is parallel to field, $U = -mB$, stable equilibrium).
(ii) Maximum when $\theta = 180^\circ$ (dipole is anti-parallel to field, $U = +mB$, unstable equilibrium).
Solution: Since they are vectors at $90^\circ$, the resultant magnetic moment $m_{net} = \sqrt{m^2 + m^2 + 2mm\cos 90^\circ} = \sqrt{2m^2} = m\sqrt{2}$.
Solution: The net magnetic force is Zero. In a uniform field, the north pole experiences a force $mB$ along the field, and the south pole experiences an equal force $mB$ opposite to the field. They cancel each other out. (Translational force is zero, though a torque may act).
Solution: $\theta_1 = 0^\circ$, $\theta_2 = 180^\circ$. $W = mB(\cos\theta_1 - \cos\theta_2) = (1.5)(0.22)(\cos 0^\circ - \cos 180^\circ) = 0.33 \times (1 - (-1)) = 0.33 \times 2 = 0.66\text{ Joules}$.
Solution: It states that the net magnetic flux through any closed surface is always zero. Mathematically, $\oint \vec{B} \cdot d\vec{S} = 0$.
Solution: It signifies that magnetic monopoles (isolated North or South poles) do not exist. Magnetic poles always exist in pairs (dipoles). The number of magnetic field lines entering a closed surface is exactly equal to the number of field lines leaving it.
Solution: If magnetic monopoles existed, the equation would be $\oint \vec{B} \cdot d\vec{S} = \mu_0 q_m$, where $q_m$ is the net enclosed magnetic pole strength.
Solution: The net magnetic flux leaving the surface is Zero. According to Gauss's Law in magnetism, the magnetic field lines that emerge from the North pole of the magnet must eventually re-enter the South pole enclosed within the same surface.
Solution: Electrostatics: $\oint \vec{E} \cdot d\vec{S} = q_{enclosed}/\epsilon_0$. Magnetism: $\oint \vec{B} \cdot d\vec{S} = 0$. The RHS is different because isolated electric charges (monopoles) exist in nature and can be enclosed by a surface, generating a net electric flux. Magnetic charges only exist in pairs, so the net enclosed magnetic charge is always zero.
Solution: Magnetic declination is the angle between the true geographic north (geographic meridian) and the magnetic north (magnetic meridian) at a given place on Earth.
Solution: The angle of dip is the angle made by the total magnetic field of the Earth ($\vec{B}$) with the horizontal direction in the magnetic meridian.
Solution: (1) Magnetic Declination, (2) Angle of Dip (Inclination), and (3) Horizontal Component of Earth's magnetic field ($B_H$).
Solution: It will orient itself perfectly vertical (pointing straight down). The angle of dip at the magnetic poles is $90^\circ$, as the Earth's magnetic field is entirely vertical there.
Solution: Given $B_H = \sqrt{3} B_V$. We know $\tan I = B_V / B_H$. $\tan I = B_V / (\sqrt{3} B_V) = 1/\sqrt{3}$. Thus, the angle of dip $I = 30^\circ$.
Solution: $B_H = B \cos I \Rightarrow B = B_H / \cos I$. Here $I = 60^\circ$ so $\cos 60^\circ = 0.5$. $B = 0.26 / 0.5 = 0.52\text{ G}$.
Solution: The angle of dip increases continuously from $0^\circ$ at the magnetic equator to $90^\circ$ at the magnetic poles.
Solution: This angle represents the Magnetic Declination at that location.
Solution: If $I=0^\circ$, then $B_V = B\sin 0^\circ = 0$. The vertical component is zero. This happens at the Magnetic Equator.
Solution: The compass north points $15^\circ$ East of True North. The ship is sailing $90^\circ$ counter-clockwise from compass north (compass West). So, $90^\circ$ CCW from $15^\circ$ East is $90^\circ - 15^\circ = 75^\circ$ West of True North. True direction is $N 75^\circ W$.
Solution: Diamagnetic materials have a small and negative susceptibility ($-1 < \chi < 0$). Paramagnetic materials have a small and positive susceptibility ($\chi > 0$).
Solution: Diamagnetic: slightly less than $1$ ($\mu_r < 1$). Paramagnetic: slightly greater than $1$ ($\mu_r > 1$). Ferromagnetic: very large compared to $1$ ($\mu_r \gg 1$).
Solution: The phenomenon of perfect exclusion of magnetic flux from a superconductor is called the Meissner effect. Since magnetic field lines are completely expelled ($B=0$ inside), $\mu_r = 0$ and $\chi = -1$. Hence they are perfect diamagnets.
Solution: Curie's law states that the magnetization of a paramagnetic material is directly proportional to the applied magnetic field intensity and inversely proportional to its absolute temperature. $\chi = C/T$ or $M = C \cdot B_0/T$ (where $C$ is Curie's constant).
Solution: Curie temperature is the specific temperature above which a ferromagnetic material loses its ferromagnetism. When heated above $T_C$, the thermal agitation disrupts the alignment of magnetic domains, and the material transitions into a paramagnetic state.
Solution: Diamagnetic: Bismuth, Copper, Water.
Paramagnetic: Aluminum.
Ferromagnetic: Iron, Nickel.
Solution: In diamagnetic atoms, electrons orbit the nucleus. According to Lenz's law, an external magnetic field induces a change in orbital motion such that the induced magnetic moment opposes the applied field, resulting in weak repulsion.
Solution: The magnetic susceptibility of a diamagnetic material is practically independent of temperature. (Thermal agitation doesn't affect the induced orbital motion).
Solution: [Student must draw: A block of material with parallel magnetic field lines approaching it, but bending outwards and moving *away* from the material, reducing the density of lines inside the material].
Solution: [Student must draw: A block of material with parallel field lines approaching it, bending sharply inwards and crowding heavily *inside* the material, leaving the exterior relatively empty].
Solution: Since the susceptibility is small and negative, the material is Diamagnetic. In a non-uniform magnetic field, it will tend to move slowly from the stronger to the weaker part of the field.
Solution: The lagging of the magnetic induction ($B$) behind the magnetizing field intensity ($H$) when a ferromagnetic material is taken through a cycle of magnetization is called hysteresis. The area under the B-H loop represents the energy dissipated as heat per unit volume per cycle of magnetization.
Solution: Retentivity is the value of the magnetic induction ($B$) left in the specimen when the magnetizing field ($H$) is reduced to zero. Coercivity is the value of the reverse magnetizing field ($H$) required to completely demagnetize the material ($B=0$).
Solution: Soft iron has high initial permeability, high retentivity (for strong fields), but crucially, low coercivity and a very narrow hysteresis loop. This means it can be magnetized and demagnetized easily with very little energy loss (heat) during AC cycles.
Solution: A permanent magnet material must have High Retentivity (to produce a strong magnetic field) and High Coercivity (so it is not easily demagnetized by stray magnetic fields, rough handling, or temperature changes).
Solution: The hysteresis loop for soft iron is narrow with a smaller area, higher retentivity, and very low coercivity. The loop for steel is wider with a larger area, slightly lower retentivity, but much higher coercivity.
Solution: Alnico (Aluminum-Nickel-Cobalt alloy) and Ticonal (Titanium-Cobalt-Nickel-Aluminum).
Solution: (i) Heating the magnet above its Curie temperature. (ii) Applying a strong alternating magnetic field and gradually reducing its amplitude to zero. (iii) Rough handling/hammering.
Solution: When an external field $H$ increases, the magnetic domains aligned with the field grow, and unaligned domains rotate to align with the field. Once all domains are perfectly aligned with the external field, the material reaches magnetic saturation. Further increase in $H$ causes no further increase in magnetization ($M$).
Solution: It should have low coercivity. When the current is switched off, the electromagnet must quickly lose its magnetism to drop the iron scrap. Low coercivity ensures easy demagnetization.
