Solution: When a magnetic dipole is placed in a uniform magnetic field $\vec{B}$ at an angle $\theta$, it experiences a restoring torque $\tau = mB\sin\theta$. Work done by external agent to rotate it against this torque by a small angle $d\theta$ is $dW = \tau d\theta = mB\sin\theta d\theta$. Total work done to rotate from $\theta_1$ to $\theta_2$ is $W = \int_{\theta_1}^{\theta_2} mB\sin\theta d\theta = mB[-\cos\theta]_{\theta_1}^{\theta_2} = -mB(\cos\theta_2 - \cos\theta_1)$. If we assume potential energy $U = 0$ when the dipole is perpendicular to the field ($\theta_1 = 90^\circ$), then energy at angle $\theta$ is $U = -mB(\cos\theta - \cos 90^\circ) = -mB\cos\theta = -\vec{m} \cdot \vec{B}$.
Solution: Initial position is parallel, $\theta_1 = 0^\circ$. Final angle $\theta_2 = 60^\circ$. Work $W = mB(\cos 0^\circ - \cos 60^\circ) = mB(1 - 0.5) = 0.5 mB$. Given $W = \sqrt{3} \text{ J}$, so $0.5 mB = \sqrt{3} \Rightarrow mB = 2\sqrt{3}$. The torque required to maintain it at $60^\circ$ is $\tau = mB\sin 60^\circ = (2\sqrt{3}) \times (\sqrt{3}/2) = 3 \text{ Nm}$.
Solution: Restoring torque on the dipole is $\tau = -mB\sin\theta$. For small angular displacements, $\sin\theta \approx \theta$, so $\tau = -mB\theta$. By Newton's second law for rotation, $\tau = I\alpha = I(d^2\theta / dt^2)$, where $I$ is moment of inertia. Thus, $I\alpha = -mB\theta \Rightarrow \alpha = -(mB/I)\theta$. This represents SHM with $\omega^2 = mB/I$. Time period $T = 2\pi/\omega = 2\pi\sqrt{I/mB}$.
Solution: $\tau = mB\sin\theta \Rightarrow B = \frac{\tau}{m\sin\theta}$. Substituting values: $B = \frac{0.024}{0.48 \times \sin 30^\circ} = \frac{0.024}{0.48 \times 0.5} = \frac{0.024}{0.24} = 0.1 \text{ Tesla}$.
Solution: Distance $d = 10 \text{ cm} = 0.1 \text{ m}$.
(a) Axial field: $B_{axial} = \frac{\mu_0}{4\pi} \frac{2m}{d^3} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} = \frac{0.96 \times 10^{-7}}{10^{-3}} = 0.96 \times 10^{-4} \text{ T} = 0.96 \text{ G}$.
(b) Equatorial field: $B_{eq} = \frac{B_{axial}}{2} = 0.48 \times 10^{-4} \text{ T} = 0.48 \text{ G}$.
Solution: Let the point be on the bisector of the angle between axes. For one magnet, the point is at an angle $\theta = 45^\circ$ from its axis. The general formula for field is $B = \frac{\mu_0 M}{4\pi d^3}\sqrt{1+3\cos^2\theta}$. For each magnet, $B_1 = B_2 = \frac{\mu_0 M}{4\pi d^3}\sqrt{1 + 3(1/\sqrt{2})^2} = \frac{\mu_0 M}{4\pi d^3}\sqrt{1 + 1.5} = \frac{\mu_0 M}{4\pi d^3}\sqrt{2.5}$. The angle between $B_1$ and $B_2$ must be calculated. Alternatively, resolve moments: $M_{net} = \sqrt{M^2+M^2} = \sqrt{2}M$ along the bisector. The point lies on the *axis* of this resultant $M_{net}$. So, $B_{net} = \frac{\mu_0}{4\pi} \frac{2M_{net}}{d^3} = \frac{\mu_0}{4\pi} \frac{2\sqrt{2}M}{d^3}$.
Solution: Magnetic moment $m = NIA = 2000 \times 4.0 \times 1.6 \times 10^{-4} = 8000 \times 1.6 \times 10^{-4} = 1.28 \text{ A m}^2$.
Torque $\tau = mB\sin\theta = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^\circ = 1.28 \times 0.075 \times 0.5 = 0.048 \text{ J}$.
Solution: $M = \text{pole strength } (m) \times L$.
(a) Transverse cut: Length becomes $L/2$, pole strength $m$ is unchanged. $M' = m \times (L/2) = M/2$.
(b) Longitudinal cut: Length $L$ unchanged, pole strength becomes $m/2$ (area is halved). $M' = (m/2) \times L = M/2$.
Solution: Consider a solenoid of length $2l$, radius $a$, with $n$ turns per unit length. Field at axial point $P$ distance $r$ from center: Consider a small element $dx$ at distance $x$ from center. Field $dB = \frac{\mu_0 (n dx) I a^2}{2[(r-x)^2 + a^2]^{3/2}}$. For $r \gg a$ and $r \gg l$, denominator $\approx r^3$. Integrating from $x = -l$ to $+l$: $B = \frac{\mu_0 n I a^2}{2r^3} \int_{-l}^{l} dx = \frac{\mu_0 n I a^2 (2l)}{2r^3}$. Multiplying num and den by $\pi$: $B = \frac{\mu_0}{4\pi} \frac{2(n \cdot 2l \cdot I \cdot \pi a^2)}{r^3} = \frac{\mu_0}{4\pi} \frac{2(NIA)}{r^3} = \frac{\mu_0}{4\pi} \frac{2m}{r^3}$, which is exactly the axial field of a bar magnet.
Solution: $\oint \vec{B} \cdot d\vec{S} = 0$. A bar magnet consists of a dipole (North and South pole of equal strength). The number of magnetic field lines originating from the North pole inside the surface is exactly equal to the number of lines terminating at the South pole inside the surface. Hence, net outgoing lines equals net incoming lines, yielding zero net flux.
Solution: The net magnetic flux is Zero. A current carrying coil acts as a magnetic dipole (with a North face and a South face). Because magnetic monopoles do not exist, any magnetic field line generated by the coil that exits the Gaussian surface must loop back and re-enter it. Therefore, $\oint \vec{B} \cdot d\vec{S} = 0$.
Solution: An electric dipole is made of two distinct, separable entities: a positive charge and a negative charge. You can enclose just one charge with a Gaussian surface and get a non-zero electric flux. A magnetic dipole cannot be separated into individual poles; cutting it just creates two smaller dipoles. You cannot enclose an isolated magnetic pole, so magnetic flux through any closed surface is always zero.
Solution: Yes, Gauss's law for magnetism ($\oint \vec{B} \cdot d\vec{S} = 0$) holds true universally for *all* magnetic fields, whether uniform or non-uniform, because it reflects the fundamental absence of magnetic monopoles anywhere in the universe.
Solution: It would indicate the discovery of isolated magnetic monopoles (a standalone North pole or South pole).
Solution: Declination ($D$): Angle between geographic and magnetic meridians. Dip ($I$): Angle Earth's total $B$-field makes with horizontal. $B_H$: Horizontal component of $B$ in magnetic meridian. Resolving $B$: $B_H = B\cos I$ and $B_V = B\sin I$. Squaring and adding: $B_H^2 + B_V^2 = B^2(\cos^2 I + \sin^2 I) = B^2$. Thus, $B = \sqrt{B_H^2 + B_V^2}$.
Solution: $B_H = B \cos I \Rightarrow B = B_H / \cos I$. $B = 0.35 / \cos(22^\circ) = 0.35 / 0.927 \approx 0.377 \text{ G}$.
Solution: $\tan I = B_V / B_H = B_V / (\sqrt{3} B_V) = 1/\sqrt{3} \Rightarrow I = 30^\circ$.
$B_H = B \cos I = 0.4 \times \cos 30^\circ = 0.4 \times (\sqrt{3}/2) = 0.2\sqrt{3} \approx 0.346 \text{ G}$.
Solution: True West is $90^\circ$ CCW from True North. The target is $15^\circ$ South of True West, so the target angle is $90^\circ + 15^\circ = 105^\circ$ CCW from True North. Magnetic North is $17^\circ$ West (CCW) of True North. Therefore, relative to Magnetic North, the target is $105^\circ - 17^\circ = 88^\circ$ CCW. The steering direction is $88^\circ$ West of Magnetic North (or $N 88^\circ W$).
Solution: At the equator, the field is purely horizontal and corresponds to the equatorial field of a dipole: $B_{eq} = \frac{\mu_0}{4\pi} \frac{m}{R^3}$. $0.4 \text{ G} = 0.4 \times 10^{-4} \text{ T}$.
$0.4 \times 10^{-4} = 10^{-7} \times \frac{m}{(6.4 \times 10^6)^3}$.
$m = \frac{0.4 \times 10^{-4} \times 262.144 \times 10^{18}}{10^{-7}} \approx 1.05 \times 10^{23} \text{ A m}^2$.
Solution: Near the geographic poles, the magnetic poles are located (which are offset). More importantly, the magnetic field is almost entirely vertical (angle of dip is close to $90^\circ$). The horizontal component $B_H$ which aligns the compass needle is extremely weak, making the compass highly unresponsive and erratic.
Solution: At the magnetic equator, the angle of dip is zero, so the vertical component of Earth's magnetic field ($B_V$) is zero. Since $B_V = 0$, the wings do not cut any vertical flux lines. Therefore, the induced emf across the wings is zero.
Solution: In the magnetic meridian, $\tan\delta = B_V / B_H$. If rotated by angle $\theta$, the effective horizontal component acting on the needle is $B_H' = B_H \cos\theta$. The vertical component $B_V$ remains unchanged. The apparent dip $\delta'$ is given by $\tan\delta' = \frac{B_V}{B_H'} = \frac{B_V}{B_H\cos\theta} = \frac{\tan\delta}{\cos\theta}$. Since $\cos\theta < 1$, $\tan\delta' > \tan\delta$, so apparent dip is always greater than true dip.
Solution:
Diamagnetic: Moves strong to weak field. $\chi$ is small, negative. $\mu_r < 1$.
Paramagnetic: Moves weak to strong field. $\chi$ is small, positive. $\mu_r > 1$.
Ferromagnetic: Moves quickly weak to strong field. $\chi$ is large, positive. $\mu_r \gg 1$.
Solution: Curie's law: The magnetic susceptibility of a paramagnetic material is inversely proportional to its absolute temperature ($\chi = C/T$). The graph is a rectangular hyperbola in the first quadrant, decreasing as $T$ increases.
Solution: Cause: Lenz's Law acting at the atomic level. External $B$ field alters the orbital speed of electrons, inducing a net magnetic moment opposing the applied field. It is independent of temperature because it is an inherent atomic property (induced orbital motion) not related to the thermal random alignment of pre-existing dipoles.
Solution: Volume of domain $V = (10^{-6})^3 = 10^{-18} \text{ m}^3 = 10^{-12} \text{ cm}^3$. Mass $= \rho V = 7.9 \times 10^{-12} \text{ g}$. Number of moles $= \frac{7.9 \times 10^{-12}}{55}$. Number of atoms $N = \frac{7.9 \times 10^{-12}}{55} \times 6.022 \times 10^{23} \approx 8.65 \times 10^{10}$ atoms. Maximum dipole moment $m_{max} = N \times m_{atom} = 8.65 \times 10^{10} \times 9.27 \times 10^{-24} \approx 8.0 \times 10^{-13} \text{ A m}^2$.
Solution: Magnetizing field $H = nI = 500 \times 0.5 = 250 \text{ A/m}$. Susceptibility $\chi = \mu_r - 1 = 1000 - 1 = 999$. Magnetization $M = \chi H = 999 \times 250 \approx 2.5 \times 10^5 \text{ A/m}$. Magnetic moment $m = M \times V = (2.5 \times 10^5) \times 10^{-4} = 25 \text{ A m}^2$.
Solution: In paramagnetic materials, individual atomic dipoles act independently, and thermal agitation largely disrupts their alignment, resulting in weak net magnetization. In ferromagnetic materials, strong quantum mechanical exchange interactions force millions of atomic dipoles to align parallel to each other within macroscopic regions called domains, leading to very strong collective magnetization.
Solution: By Curie's Law, $\chi \propto 1/T \Rightarrow \chi_1 T_1 = \chi_2 T_2$. $(1.2 \times 10^{-5}) \times 300 = (1.8 \times 10^{-5}) \times T_2$. $T_2 = \frac{1.2 \times 300}{1.8} = \frac{360}{1.8} = 200\text{ K}$.
Solution: [Student must draw an S-shaped loop]. Retentivity is the positive y-intercept (value of B when H is 0). Coercivity is the negative x-intercept (value of reverse H required to make B zero).
Solution: Transformers and AC electromagnets operate on alternating currents, subjecting the core to hundreds of magnetization-demagnetization cycles per second. Soft iron has a much narrower hysteresis loop area compared to steel, meaning it dissipates far less energy (heat) per cycle, preventing overheating and improving efficiency.
Solution: The area enclosed by the B-H loop represents the energy dissipated as heat per unit volume of the material during one complete cycle of magnetization and demagnetization.
Solution: Energy loss per cycle for the entire core = Area $\times$ Volume = $100 \text{ J/m}^3 \times 0.01 \text{ m}^3 = 1\text{ Joule}$. Since frequency is $50\text{ Hz}$ (50 cycles per second), Power loss = Energy/cycle $\times$ frequency = $1\text{ J} \times 50\text{ s}^{-1} = 50\text{ Watts}$.
Solution: Coercivity is the demagnetizing field $H_c$. For a solenoid, $H = nI = (N/L)I$. $4 \times 10^3 = \frac{500}{0.1} \times I = 5000 I$. Therefore, $I = \frac{4000}{5000} = 0.8\text{ A}$.
Solution: Permanent Magnet: Requires High Retentivity (strong field) and High Coercivity (hard to demagnetize). Example: Steel, Alnico.
Electromagnet: Requires High Permeability (strong temporary field), Low Retentivity (loses magnetism quickly when current stops), and Low Coercivity. Example: Soft Iron.
Solution: Dropping (mechanical shock) or heating provides kinetic/thermal energy to the atoms. This energy violently disrupts the strict parallel alignment of the magnetic domains. The domains randomize their orientations, causing the net macroscopic magnetic field to cancel out and disappear.
Solution: No. While high retentivity provides a strong initial magnetic field, low coercivity means the material is highly vulnerable. Minor stray magnetic fields, slight temperature variations, or mild mechanical shocks will easily destroy its magnetization, making it unreliable as a "permanent" magnet.
Solution: Alnico is an alloy of Aluminum, Nickel, Cobalt, Copper, and Iron. It is suited for permanent magnets because it possesses exceptionally high coercivity (making it virtually immune to accidental demagnetization) and reasonably high retentivity.
