Solution: Initial moment $M = p \cdot L$, where $p$ is pole strength and $L$ is length. When bent into a semicircle of radius $R$, the length $L = \pi R \Rightarrow R = L/\pi$. The new distance between the poles is the diameter, $2R = 2L/\pi$. The pole strength $p$ remains unchanged. New moment $M' = p \cdot (2L/\pi) = \frac{2}{\pi}(pL) = \frac{2M}{\pi}$.
Solution: Initial $M = p \cdot L$. Arc subtends $60^\circ = \pi/3$ radians. $L = R\theta = R(\pi/3) \Rightarrow R = 3L/\pi$. The chord length (new distance between poles) forms an equilateral triangle with the two radii (since angle is $60^\circ$ and two sides are $R$). So, chord $d = R = 3L/\pi$. New moment $M' = p \cdot d = p(3L/\pi) = \frac{3M}{\pi}$.
Solution: The magnetic field due to $m_1$ at the location of $m_2$ is an axial field: $B_1 = \frac{\mu_0}{4\pi} \frac{2m_1}{r^3}$. The potential energy of $m_2$ in this field is $U = -\vec{m}_2 \cdot \vec{B}_1$. Since they are coaxial, $U = -m_2 B_1 = -m_2 \left(\frac{\mu_0}{4\pi} \frac{2m_1}{r^3}\right) = -\frac{\mu_0}{4\pi} \frac{2m_1 m_2}{r^3}$. The force is the negative gradient of potential energy: $F = -\frac{dU}{dr} = -\frac{d}{dr}\left(-\frac{\mu_0}{4\pi} \frac{2m_1 m_2}{r^3}\right) = \frac{\mu_0}{4\pi}(2m_1 m_2)(-3r^{-4}) = -\frac{\mu_0}{4\pi} \frac{6m_1 m_2}{r^4}$. The negative sign indicates an attractive force (if aligned parallel).
Solution: Initial time period $T = 2\pi\sqrt{I/MB}$. Let mass be $W$, length $L$, width $b$. $I = W(L^2+b^2)/12$. When cut into 4 equal pieces, the mass of one piece $W' = W/4$. Length $L' = L/2$, width $b' = b/2$. New moment of inertia $I' = \frac{W'((L')^2+(b')^2)}{12} = \frac{W/4(L^2/4 + b^2/4)}{12} = \frac{1}{16} \frac{W(L^2+b^2)}{12} = \frac{I}{16}$. Magnetic moment $M = \text{pole strength} \times \text{length}$. Pole strength is halved (cut along length), length is halved (cut along width). $M' = (p/2) \times (L/2) = M/4$. New time period $T' = 2\pi\sqrt{\frac{I/16}{(M/4)B}} = 2\pi\sqrt{\frac{I}{4MB}} = \frac{1}{2} \left(2\pi\sqrt{\frac{I}{MB}}\right) = \frac{T}{2}$.
Solution: Point $P(a, a)$ is at distance $r = \sqrt{a^2+a^2} = a\sqrt{2}$. The angle $\theta$ with the x-axis is $45^\circ$. Radial component $B_r = \frac{\mu_0}{4\pi} \frac{2m\cos\theta}{r^3} = \frac{\mu_0}{4\pi} \frac{2m_0(1/\sqrt{2})}{(a\sqrt{2})^3} = \frac{\mu_0}{4\pi} \frac{\sqrt{2}m_0}{2\sqrt{2}a^3} = \frac{\mu_0 m_0}{8\pi a^3}$. Transverse component $B_\theta = \frac{\mu_0}{4\pi} \frac{m\sin\theta}{r^3} = \frac{\mu_0}{4\pi} \frac{m_0(1/\sqrt{2})}{2\sqrt{2}a^3} = \frac{\mu_0 m_0}{16\pi a^3}$. Magnitude $|\vec{B}| = \sqrt{B_r^2 + B_\theta^2} = \frac{\mu_0 m_0}{16\pi a^3}\sqrt{4+1} = \frac{\sqrt{5}\mu_0 m_0}{16\pi a^3}$.
Solution: At the initial neutral point, the magnet's equatorial field exactly cancels Earth's horizontal field: $B_{eq} = B_H \Rightarrow \frac{\mu_0}{4\pi}\frac{m}{d^3} = B_H$. When rotated by $90^\circ$, the point $d$ now lies on the *axial* line of the magnet. The magnet's field there is $B_{axial} = \frac{\mu_0}{4\pi}\frac{2m}{d^3} = 2 B_{eq} = 2 B_H$. This field is perpendicular to Earth's $B_H$ (since the magnet was rotated $90^\circ$). The net field is the vector sum of two perpendicular fields: $B_{net} = \sqrt{B_{axial}^2 + B_H^2} = \sqrt{(2B_H)^2 + B_H^2} = \sqrt{5 B_H^2} = \sqrt{5} B_H$.
Solution: Time period $T \propto 1/\sqrt{B_{net}}$. Initially, $B_{net1} = B_H = 24 \mu\text{T}$, $T_1 = 2\text{ s}$. When opposing field is added, $B_{net2} = 24 - 18 = 6 \mu\text{T}$. $\frac{T_2}{T_1} = \sqrt{\frac{B_{net1}}{B_{net2}}} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$. Therefore, new time period $T_2 = 2 \times T_1 = 2 \times 2 = 4\text{ s}$.
Solution: Magnetic moment of the coil $m_{coil} = NIA = 10 \times 2 \times 10^{-4} = 20 \times 10^{-4} = 2 \times 10^{-3} \text{ A m}^2$. The field produced by the bar magnet at $d=1\text{ m}$ on its axial line is $B = \frac{\mu_0}{4\pi}\frac{2M}{d^3} = 10^{-7} \times \frac{2 \times 10}{1^3} = 2 \times 10^{-6}\text{ T}$. Since the coil's axis is perpendicular to the magnet's axis, the magnetic moment of the coil is perpendicular to $B$. Torque $\tau = m_{coil} B \sin 90^\circ = (2 \times 10^{-3}) \times (2 \times 10^{-6}) \times 1 = 4 \times 10^{-9}\text{ N m}$.
Solution: Time period $T = 6.70 / 10 = 0.67\text{ s}$. $T = 2\pi\sqrt{I/MB} \Rightarrow T^2 = 4\pi^2 I / (MB) \Rightarrow B = \frac{4\pi^2 I}{M T^2}$. $B = \frac{4 \times (3.14)^2 \times 8.5 \times 10^{-7}}{5 \times 10^{-2} \times (0.67)^2} \approx \frac{39.44 \times 8.5 \times 10^{-7}}{0.05 \times 0.4489} \approx \frac{335.24 \times 10^{-7}}{0.0224} \approx 1.5 \times 10^{-3}\text{ Tesla}$.
Solution: Let the mid-point be at distance $r = d/2$ from both. For magnet 1 (horizontal), the mid-point lies on its axial line. Field $B_1 = \frac{\mu_0}{4\pi} \frac{2M}{(d/2)^3} = \frac{\mu_0}{4\pi} \frac{16M}{d^3}$ (horizontal). For magnet 2 (vertical), the mid-point lies on its equatorial line. Field $B_2 = \frac{\mu_0}{4\pi} \frac{M}{(d/2)^3} = \frac{\mu_0}{4\pi} \frac{8M}{d^3}$ (also horizontal, but check directions). Wait, if they form a "T" shape, $B_2$ is parallel to magnet 1's axis. So $B_1$ and $B_2$ are perpendicular to each other. $B_{net} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{4\pi d^3} \sqrt{(16M)^2 + (8M)^2} = \frac{\mu_0 M}{4\pi d^3} \sqrt{256 + 64} = \frac{\mu_0 M}{4\pi d^3} \sqrt{320} = \frac{\mu_0 M}{4\pi d^3} 8\sqrt{5}$.
Solution: Let true dip be $\delta$, so $\tan\delta = B_V / B_H$. In plane 1 (angle $\theta$ with meridian), $B_{H1} = B_H \cos\theta$, so $\tan\delta_1 = B_V / (B_H \cos\theta) \Rightarrow \cot\delta_1 = \frac{B_H \cos\theta}{B_V}$. In plane 2 (angle $90^\circ+\theta$), $B_{H2} = B_H \cos(90^\circ+\theta) = B_H \sin\theta$, so $\tan\delta_2 = B_V / (B_H \sin\theta) \Rightarrow \cot\delta_2 = \frac{B_H \sin\theta}{B_V}$. Squaring and adding: $\cot^2\delta_1 + \cot^2\delta_2 = \frac{B_H^2 \cos^2\theta}{B_V^2} + \frac{B_H^2 \sin^2\theta}{B_V^2} = \frac{B_H^2}{B_V^2}(\cos^2\theta + \sin^2\theta) = \left(\frac{B_H}{B_V}\right)^2 = \cot^2\delta$. Proved.
Solution: $\tan\delta' = \frac{\tan\delta}{\cos\theta} = \frac{\tan 30^\circ}{\cos 45^\circ} = \frac{1/\sqrt{3}}{1/\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{2/3}$. Therefore, $\delta' = \tan^{-1}(\sqrt{2/3})$.
Solution: At geomagnetic latitude $\lambda$, the colatitude (angle from pole) is $\theta = 90^\circ - \lambda$. The Earth's magnetic dipole $M$ is at the center. $B_V$ is the radial component: $B_r = \frac{\mu_0}{4\pi} \frac{2M\cos\theta}{R^3} = \frac{\mu_0}{4\pi} \frac{2M\sin\lambda}{R^3}$. $B_H$ is the transverse component: $B_\theta = \frac{\mu_0}{4\pi} \frac{M\sin\theta}{R^3} = \frac{\mu_0}{4\pi} \frac{M\cos\lambda}{R^3}$. Total $B = \sqrt{B_r^2 + B_\theta^2} = \frac{\mu_0 M}{4\pi R^3}\sqrt{4\sin^2\lambda + \cos^2\lambda} = \frac{\mu_0 M}{4\pi R^3}\sqrt{1 + 3\sin^2\lambda}$.
Solution: At the equator ($\lambda = 0^\circ$), $B_{eq} = B_H = \frac{\mu_0 M}{4\pi R^3} = 0.4 \text{ G}$. At latitude $\lambda = 30^\circ$, from the derivation above, $B = B_{eq} \sqrt{1 + 3\sin^2\lambda} = 0.4 \sqrt{1 + 3\sin^2 30^\circ} = 0.4 \sqrt{1 + 3(1/4)} = 0.4 \sqrt{1 + 0.75} = 0.4 \sqrt{1.75} = 0.4 \sqrt{7/4} = 0.2\sqrt{7} \approx 0.2 \times 2.645 = 0.529 \text{ G}$.
Solution: Initially $B_H = B\cos 45^\circ = 0.6 \times (1/\sqrt{2}) = 0.3\sqrt{2} \text{ G}$. New total field $B' = 1.1 \times 0.6 = 0.66 \text{ G}$. Since $B_H$ is constant, $B_H = B' \cos\delta' \Rightarrow 0.3\sqrt{2} = 0.66 \cos\delta'$. $\cos\delta' = \frac{0.3\sqrt{2}}{0.66} = \frac{30\sqrt{2}}{66} = \frac{5\sqrt{2}}{11} \approx \frac{5 \times 1.414}{11} = \frac{7.07}{11} \approx 0.642$. $\delta' = \cos^{-1}(0.642) \approx 50^\circ$.
Solution: From dipole model, $\tan\delta = \frac{B_V}{B_H} = \frac{2\sin\lambda}{\cos\lambda} = 2\tan\lambda$. Differentiating w.r.t $\lambda$: $\sec^2\delta \frac{d\delta}{d\lambda} = 2\sec^2\lambda$. Using $\sec^2\delta = 1 + \tan^2\delta = 1 + 4\tan^2\lambda$. Rate $\frac{d\delta}{d\lambda} = \frac{2\sec^2\lambda}{1 + 4\tan^2\lambda} = \frac{2/\cos^2\lambda}{1 + 4\sin^2\lambda/\cos^2\lambda} = \frac{2}{\cos^2\lambda + 4\sin^2\lambda} = \frac{2}{1 + 3\sin^2\lambda}$.
Solution: In a tangent galvanometer, $I = K\tan\theta$, where $K$ is the reduction factor ($K \propto B_H$). $I = K\tan 60^\circ = K\sqrt{3}$. For current $I/3$: $I/3 = K\tan\theta'$. Substituting $I$: $\frac{K\sqrt{3}}{3} = K\tan\theta' \Rightarrow \tan\theta' = \frac{1}{\sqrt{3}} \Rightarrow \theta' = 30^\circ$.
Solution: If magnet North points South, the magnetic field lines of the magnet align against $B_H$ on the *axial* line. Neutral points are on the axial line. $B_{axial} = B_H \Rightarrow \frac{\mu_0}{4\pi}\frac{2M}{d^3} = B_H$. $10^{-7} \times \frac{2M}{(0.3)^3} = 0.34 \times 10^{-4}$. $\frac{2M}{0.027} = 340 \Rightarrow 2M = 9.18 \Rightarrow M = 4.59 \text{ A m}^2$.
Solution: Now magnet lines oppose Earth's field on the *equatorial* line. $B_{eq} = B_H \Rightarrow \frac{\mu_0}{4\pi}\frac{M}{r^3} = B_H$. We know $\frac{\mu_0}{4\pi}\frac{2M}{d^3} = B_H$. Equating them: $\frac{M}{r^3} = \frac{2M}{d^3} \Rightarrow r^3 = d^3 / 2 \Rightarrow r = d / 2^{1/3} = 30 / 1.26 \approx 23.8 \text{ cm}$ on the equatorial line.
Solution: Magnetic force provides centripetal force: $ev B_V = \frac{mv^2}{R} \Rightarrow v = \frac{e B_V R}{m}$. The kinetic energy is gained from accelerating voltage $V$: $eV = \frac{1}{2}mv^2$. Substitute $v$: $eV = \frac{1}{2}m \left(\frac{e B_V R}{m}\right)^2 = \frac{e^2 B_V^2 R^2}{2m}$. Therefore, $V = \frac{e B_V^2 R^2}{2m}$.
Solution: Mean circumference $L = 2\pi R = 2\pi(0.15) = 0.3\pi \text{ m}$. Number of turns per unit length $n = N/L = 500 / 0.3\pi$. Magnetic Intensity $H = nI = \frac{500}{0.3\pi} \times 2 = \frac{1000}{0.3\pi} \approx 1061 \text{ A/m}$. $B = \mu_0 \mu_r H = (4\pi \times 10^{-7}) \times 1500 \times \frac{1000}{0.3\pi} = 4 \times 10^{-7} \times 1500 \times \frac{1000}{0.3} = 4 \times 1500 \times 3333 \times 10^{-7} = 2 \text{ T}$. Magnetization $M \approx \mu_r H$ (since $\mu_r \gg 1$) $= 1500 \times 1061 \approx 1.59 \times 10^6 \text{ A/m}$.
Solution: Curie's Law $\chi \propto 1/T$. $\chi_1 T_1 = \chi_2 T_2$.
At 200 K: $\chi_{200} = \chi_{300} \times \frac{300}{200} = 1.2 \times 10^{-5} \times 1.5 = 1.8 \times 10^{-5}$.
At 600 K: $\chi_{600} = \chi_{300} \times \frac{300}{600} = 1.2 \times 10^{-5} \times 0.5 = 0.6 \times 10^{-5}$.
Solution: Volume $V = 10^{-18} \text{ m}^3 = 10^{-12} \text{ cm}^3$. Mass $m = 7.9 \times 10^{-12} \text{ g}$. Atoms $N = \frac{7.9 \times 10^{-12}}{55} \times 6.02 \times 10^{23} = 8.65 \times 10^{10}$. Max Moment $m_{max} = N \mu = 8.65 \times 10^{10} \times 9.27 \times 10^{-24} = 8 \times 10^{-13} \text{ A m}^2$. Magnetization $M_{max} = m_{max}/V = \frac{8 \times 10^{-13}}{10^{-18}} = 8 \times 10^5 \text{ A/m}$.
Solution: Meissner effect perfectly expels flux, so Internal $B = 0$. Since $B = \mu_r B_{ext}$, $\mu_r = 0$. Also $\chi = \mu_r - 1 = 0 - 1 = -1$ (Perfect diamagnet). Because it perfectly excludes magnetic fields, it experiences a maximum repulsive force when placed over a magnet, allowing it to levitate stably.
Solution: $B = \mu_0(H+M) \Rightarrow M = \frac{B}{\mu_0} - H$. $\frac{B}{\mu_0} = \frac{3.14}{4 \times 3.14 \times 10^{-7}} = \frac{1}{4 \times 10^{-7}} = 2.5 \times 10^6 \text{ A/m}$. $M = 2.5 \times 10^6 - 0.002 \times 10^6 \approx 2.5 \times 10^6 \text{ A/m}$. Pole strength $p = M \times \text{Area}$. Area $A = 0.5 \times 0.2 \text{ cm}^2 = 0.1 \times 10^{-4} \text{ m}^2 = 10^{-5} \text{ m}^2$. $p = 2.5 \times 10^6 \times 10^{-5} = 25 \text{ A m}$.
Solution: By definition, Magnetization $\vec{M} = \chi \vec{H}$. Substitute into the equation: $\vec{B} = \mu_0(\vec{H} + \chi \vec{H}) = \mu_0(1+\chi)\vec{H}$. Also by definition, $\vec{B} = \mu\vec{H} = \mu_0\mu_r\vec{H}$. Comparing the two, $\mu_0\mu_r\vec{H} = \mu_0(1+\chi)\vec{H}$, thus $\mu_r = 1 + \chi$.
Solution: Paramagnetic liquid moves from weak to strong field, so the level rises in the arm between the poles. The magnetic energy density pulls the liquid up. Change in pressure $\Delta P = \frac{1}{2} M B = \frac{1}{2} \chi H \cdot (\mu_0 H) = \frac{1}{2}\mu_0 \chi H^2$. Equating to hydrostatic pressure $\rho g h$: $h = \frac{\mu_0 \chi H^2}{2\rho g}$ or $\frac{\chi B^2}{2\mu_0 \rho g}$.
Solution: Using classical statistical mechanics (Langevin theory), $M = M_{sat} (\frac{\mu B}{k_B T})$ for small $\mu B/kT$. However, for $50\%$ saturation, we cannot use the linear approximation. The exact function is $M = M_{sat} \tanh(\frac{\mu B}{k_B T})$. So $\tanh(\frac{\mu B}{k_B T}) = 0.5 \Rightarrow \frac{\mu B}{k_B T} \approx 0.55$. $T = \frac{\mu B}{0.55 k_B} = \frac{10^{-23} \times 2}{0.55 \times 1.38 \times 10^{-23}} = \frac{2}{0.759} \approx 2.6 \text{ K}$. (Only achievable at cryogenic temperatures).
Solution: Above the Curie temperature $T_C$, ferromagnetic materials lose long-range order and become paramagnetic. Their susceptibility does not follow the simple Curie's Law ($\chi = C/T$) but follows the Curie-Weiss law: $\chi = \frac{C}{T - T_C}$ (for $T > T_C$). This accounts for the residual internal molecular fields aiding magnetization.
Solution: Magnetic induction $B = \mu H = (8 \times 10^{-3}) \times 160 = 1.28 \text{ T}$. We know $B = \mu_0(H+M) \Rightarrow M = \frac{B}{\mu_0} - H$. $M = \frac{1.28}{4\pi \times 10^{-7}} - 160 = \frac{1.28}{12.56 \times 10^{-7}} - 160 \approx 1.019 \times 10^6 - 160 \approx 1.02 \times 10^6 \text{ A/m}$.
Solution: Demagnetizing field needed is $H_c = 3 \times 10^3 \text{ A/m}$. For a solenoid, $H = nI = (N/L)I$. $3 \times 10^3 = \frac{500}{0.1} I = 5000 I$. $I = 3000 / 5000 = 0.6 \text{ A}$.
Solution: Energy loss per cycle = Area $\times$ Volume = $0.1 \text{ J/m}^3 \times 10^{-3} \text{ m}^3 = 10^{-4} \text{ J}$. Power loss = Energy per cycle $\times$ frequency = $10^{-4} \times 50 = 5 \times 10^{-3} \text{ W} = 5 \text{ mW}$.
Solution: Power supplied to a solenoid core is $P = VI$. Induced emf $V = N A \frac{dB}{dt}$. $P = (N A \frac{dB}{dt}) I$. Work done $dW = P dt = NA I dB$. We know $H = nI = (N/l)I \Rightarrow NI = Hl$. $dW = (Hl) A dB = H (Al) dB = H V_{ol} dB$. Work per unit volume $dW/V_{ol} = H dB$. Over a full cycle, total work per unit volume $W = \oint H \, dB$, which is the area of the loop.
Solution: (i) Permanent Magnets need high field and resistance to demagnetization: Material A. (ii) Electromagnets need to demagnetize easily and not retain field: Material B (though soft iron actually has high retentivity, the "low C" is critical, making C also acceptable, but B is standard text-book answer for AC cores). (iii) Magnetic tapes need to retain memory but be writable: Material C (High retentivity to store data, moderate/low coercivity to be overwritten).
Solution: Heat produced per second per $m^3$ = Area $\times$ frequency = $300 \times 50 = 15000 \text{ J/(s m}^3)$. Heat $Q = ms\Delta T = (\rho V) s \Delta T$. Rate of heat $P = \rho V s (dT/dt)$. Power per unit volume $P/V = \rho s (dT/dt)$. $15000 = 7500 \times 400 \times (dT/dt) \Rightarrow dT/dt = \frac{15000}{3 \times 10^6} = 5 \times 10^{-3} \text{ K/s} = 0.005 \text{ K/s}$.
Solution: Cores are laminated to reduce *Eddy Current* losses. Changing magnetic flux induces circular currents (eddy currents) in bulk metal, causing ohmic heating ($I^2R$). Lamination breaks the electrical circuit. This is entirely separate from Hysteresis loss, which is the energy spent forcing magnetic domains to flip back and forth (narrow loop minimizes this).
Solution: To bring flux to zero, we must apply a reverse field equal to the coercivity $H_c = 5000 \text{ A/m}$. For a toroid, $H = \frac{NI}{2\pi r}$. $5000 = \frac{1000 \times I}{2\pi(0.1)} \Rightarrow 5000 = \frac{1000 I}{0.628} \Rightarrow I = \frac{5000 \times 0.628}{1000} = 5 \times 0.628 = 3.14 \text{ A}$.
Solution: At $0 \text{ K}$, thermal agitation is zero. Exchange coupling perfectly aligns all atomic dipoles within a domain (100% saturation). As $T$ increases, thermal energy competes with exchange coupling, causing slight misalignments and reducing net magnetization. At $T = T_C$, thermal energy completely overwhelms the exchange interaction. Domains break apart, long-range order is destroyed, and the material transitions into a paramagnetic state where dipoles act independently.
Solution: A permanent magnet's maximum field is strictly limited by the intrinsic retentivity of the material. To achieve a very strong macroscopic field, you need a massive volume of the material. An electromagnet's field ($B = \mu_0 \mu_r n I$) can be amplified by increasing the current $I$ and turns $n$. Thus, an electromagnet can achieve extremely high magnetic fields with a significantly smaller mass and volume of iron core compared to a permanent magnet.
Solution: The internal $\vec{B}$ field is generally not uniform, especially near the ends. Magnetization creates "fictitious" magnetic poles at the surfaces, which generate an internal field $\vec{H}_d$ opposing the magnetization, called the Demagnetizing Field. For a short, flat cylinder (coin shape), the poles are close together, resulting in a very strong demagnetizing field. For a long, thin needle, the poles are far apart, making the internal demagnetizing field very weak, ensuring a more uniform and stable internal $B$ field.
