How to Use This Problem Set

This is your full workout for Magnetism and Matter, grouped by theme: magnetic moments and bar-magnet fields, torque and energy of a dipole, oscillations and two-dipole configurations, Gauss's law reasoning, and the B-H-M-χ\chi-μr\mu_r toolbox.

Keep these handy:

  • m=NIAm = NIA; axial BA=μ04π2mr3B_A = \frac{\mu_0}{4\pi}\frac{2m}{r^3}; equatorial BE=μ04πmr3B_E = \frac{\mu_0}{4\pi}\frac{m}{r^3} in magnitude, directed opposite to m⃗\vec{m}
  • τ⃗=m⃗×B⃗\vec{\tau} = \vec{m}\times\vec{B}; U=−m⃗⋅B⃗U = -\vec{m}\cdot\vec{B}; W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos\theta_1 - \cos\theta_2)
  • T=2πI/(mB)T = 2\pi\sqrt{\mathcal{I}/(mB)} (oscillating needle)
  • Gauss: ∑B⃗⋅ΔS⃗=0\sum \vec{B}\cdot\Delta\vec{S} = 0 for any closed surface
  • H=nIH = nI; M=χHM = \chi H; B=μ0(H+M)=μ0μrHB = \mu_0(H+M) = \mu_0\mu_r H; μr=1+χ\mu_r = 1+\chi
  • μ04π=10−7\frac{\mu_0}{4\pi} = 10^{-7} T m/A; 1 G =10−4= 10^{-4} T

Work with units throughout, and try each problem before reading its solution.

Solved Examples - Magnetic Moments & Bar-Magnet Fields

Example 1. A coil of 800 turns, area 2.5×10−42.5 \times 10^{-4} m2^2, carries 3.0 A. Its magnetic moment?

Solution: m=NIA=800×3.0×2.5×10−4=0.6m = NIA = 800 \times 3.0 \times 2.5 \times 10^{-4} = 0.6 A m2^2.

Example 2. Axial field of a short magnet (m=0.6m = 0.6 A m2^2) at 10 cm:

Solution: BA=10−7×2×0.6(0.1)3=10−7×1200=1.2×10−4B_A = 10^{-7} \times \frac{2 \times 0.6}{(0.1)^3} = 10^{-7} \times 1200 = 1.2 \times 10^{-4} T, along m⃗\vec{m}.

Example 3. Equatorial field of the same magnet at the same distance:

Solution: Half the axial value in magnitude: BE=0.6×10−4B_E = 0.6 \times 10^{-4} T, directed opposite to m⃗\vec{m}.

Example 4. A magnet of moment 1.2 A m2^2 is cut transverse to its length into two equal pieces. Moment of each?

Solution: Like halving a solenoid's turns: m′=1.2/2=0.6m' = 1.2/2 = 0.6 A m2^2 each. For a lengthwise cut into two equal parts, the area is halved and the moment is also halved.

Example 5. Where on the axis is the field of a magnet (m=0.4m = 0.4 A m2^2) equal to 0.8×10−40.8 \times 10^{-4} T?

Solution: r3=10−7×2mB=10−7×0.80.8×10−4=10−3r^3 = 10^{-7}\times\frac{2m}{B} = 10^{-7} \times \frac{0.8}{0.8 \times 10^{-4}} = 10^{-3}, so r=0.1r = 0.1 m.

Example 6. Ratio of magnitudes of axial and equatorial fields at the same distance from a short magnet:

Solution: ∣BA∣/∣BE∣=2|B_A|/|B_E| = 2, always in the far zone. The equatorial field is opposite to m⃗\vec{m}, so this ratio compares magnitudes.

Example 7. A solenoid of 400 turns, area 4×10−44 \times 10^{-4} m2^2, current 5.0 A. Field on its axis at 0.2 m (far zone)?

Solution: m=400×5.0×4×10−4=0.8m = 400 \times 5.0 \times 4\times10^{-4} = 0.8 A m2^2. B=10−7×2×0.8(0.2)3=10−7×200=2×10−5B = 10^{-7}\times\frac{2 \times 0.8}{(0.2)^3} = 10^{-7} \times 200 = 2 \times 10^{-5} T.

Example 8. Field of a short magnet (m=1.0m = 1.0 A m2^2) at 10 cm, 60 degrees from the axis [JEE pattern]:

Solution: B=μ0m4πr31+3cos⁡260∘=10−41.75≈1.32×10−4B = \frac{\mu_0 m}{4\pi r^3}\sqrt{1+3\cos^2 60^{\circ}} = 10^{-4}\sqrt{1.75} \approx 1.32 \times 10^{-4} T.

Solved Examples - Torque & Energy of a Dipole

Example 9. Torque on a magnet (m=0.6m = 0.6 A m2^2) at 30 degrees to a 0.2 T field:

Solution: τ=mBsin⁡θ=0.6×0.2×0.5=0.06\tau = mB\sin\theta = 0.6 \times 0.2 \times 0.5 = 0.06 N m.

Example 10. Maximum torque on the same magnet in the same field:

Solution: τmax=mB=0.12\tau_{max} = mB = 0.12 N m, at θ=90∘\theta = 90^{\circ}.

Example 11. Potential energy of the same magnet when aligned with the field:

Solution: U=−mB=−0.12U = -mB = -0.12 J (most stable orientation).

Example 12. Work to flip the magnet (m=0.6m = 0.6, B=0.2B = 0.2 T) from aligned to anti-aligned:

Solution: W=2mB=0.24W = 2mB = 0.24 J.

Example 13. Work to rotate it from 60 degrees to 90 degrees:

Solution: W=mB(cos⁡60∘−cos⁡90∘)=0.12×0.5=0.06W = mB(\cos 60^{\circ} - \cos 90^{\circ}) = 0.12 \times 0.5 = 0.06 J.

Example 14. A magnet (m=0.32m = 0.32 J/T) in a 0.15 T field: energies of stable and unstable equilibrium?

Solution: U=∓mB=∓0.048U = \mp mB = \mp 0.048 J (−- at θ=0\theta = 0 stable, ++ at 180∘180^{\circ} unstable); torque is zero at both.

Example 15. Where is U=0U = 0 for a dipole, and what is the torque there?

Solution: At θ=90∘\theta = 90^{\circ} (the chosen zero); torque there is maximum, mBmB. Energy zero does not mean equilibrium!

Example 16. A dipole swings freely from 90 degrees to alignment. Kinetic energy gained (m=0.6m = 0.6, B=0.2B = 0.2 T)?

Solution: KE =U(90∘)−U(0)=0−(−mB)=0.12= U(90^{\circ}) - U(0) = 0 - (-mB) = 0.12 J.

Solved Examples - Oscillations & Two-Dipole Configurations

Example 17. A needle: m=0.04m = 0.04 A m2^2, I=1.6×10−5\mathcal{I} = 1.6 \times 10^{-5} kg m2^2, B=0.01B = 0.01 T. Period of small oscillations?

Solution: mB=4×10−4mB = 4 \times 10^{-4}; T=2π1.6×10−54×10−4=2π0.04=2π(0.2)≈1.26T = 2\pi\sqrt{\frac{1.6\times10^{-5}}{4\times10^{-4}}} = 2\pi\sqrt{0.04} = 2\pi(0.2) \approx 1.26 s.

Example 18. The field is quadrupled. New period?

Solution: T∝1/BT \propto 1/\sqrt{B}: T′=T/2≈0.63T' = T/2 \approx 0.63 s.

Example 19. Dipole Q sits on the axis of dipole P. For stable equilibrium, m⃗Q\vec{m}_Q must be:

Solution: Parallel to B⃗P\vec{B}_P, which on the axis is along m⃗P\vec{m}_P — so parallel to m⃗P\vec{m}_P (on the normal bisector it would be antiparallel).

Example 20. Two magnets (m1=m2=2.0m_1 = m_2 = 2.0 A m2^2) are coaxial and aligned, 0.5 m apart. Interaction energy?

Solution: B1=10−7×2×2.0(0.5)3=3.2×10−6B_1 = 10^{-7}\times\frac{2\times2.0}{(0.5)^3} = 3.2\times10^{-6} T; U=−m2B1=−6.4×10−6U = -m_2 B_1 = -6.4 \times 10^{-6} J (bound, stable).

Example 21. Among all positions/orientations of Q at distance r from P, the minimum-energy one is:

Solution: Q on P's axis with moments parallel: U=−μ02mPmQ4πr3U = -\frac{\mu_0 2 m_P m_Q}{4\pi r^3} — the axial field is double the equatorial field in magnitude.

Solved Examples - Gauss's Law of Magnetism

Example 22. Net magnetic flux through any closed surface containing a full bar magnet?

Solution: Zero — every line exiting re-enters; no monopoles, ∑B⃗⋅ΔS⃗=0\sum\vec{B}\cdot\Delta\vec{S} = 0.

Example 23. A closed surface encloses just the S-pole end of a magnet. Net flux?

Solution: Still zero: external lines entering near S are balanced by lines leaving through the magnet's internal cross-section.

Example 24. If a monopole of magnetic charge qmq_m were enclosed, the flux would be:

Solution: ∮B⃗⋅dS⃗=μ0qm\oint\vec{B}\cdot d\vec{S} = \mu_0 q_m — the hypothetical modification of Gauss's law.

Example 25. A 0.3 T uniform field is parallel to one edge set of a cube of side 0.2 m. Flux through each face and net flux?

Solution: The two faces whose outward normals are parallel/antiparallel to B⃗\vec{B} have flux ϕ=±BA=±(0.3×0.04)=±1.2×10−2\phi = \pm BA = \pm(0.3 \times 0.04) = \pm 1.2\times10^{-2} Wb. The four faces whose planes are parallel to B⃗\vec{B} have zero flux. Hence the net flux is zero.

Example 26. A diagram shows field lines radiating straight out of a point. Magnetic or not?

Solution: Not a physical magnetic field pattern in ordinary magnetostatics — non-zero net flux from an isolated point would mean a magnetic monopole. In standard board-level physics, such a purely radial point pattern represents an electrostatic field of a point charge, not the field of a bar magnet.

Solved Examples - The B, H, M, chi, mu_r Toolbox

Example 27. A solenoid has 500 turns/m and carries 3.0 A. H inside?

Solution: H=nI=1500H = nI = 1500 A/m — independent of any core.

Example 28. A core of μr=600\mu_r = 600 fills it. B inside?

Solution: B=μ0μrH=4π×10−7×600×1500≈1.13B = \mu_0\mu_r H = 4\pi\times10^{-7} \times 600 \times 1500 \approx 1.13 T.

Example 29. Magnetisation of that core?

Solution: M=(μr−1)H=599×1500≈9.0×105M = (\mu_r - 1)H = 599 \times 1500 \approx 9.0 \times 10^5 A/m.

Example 30. Bismuth has χ=−1.6×10−4\chi = -1.6 \times 10^{-4}. For H=105H = 10^5 A/m, find M.

Solution: M=χH=−16M = \chi H = -16 A/m — opposite to H (diamagnetic).

Example 31. Its permeability?

Solution: μ=μ0(1+χ)=4π×10−7(1−1.6×10−4)≈1.256×10−6\mu = \mu_0(1+\chi) = 4\pi\times10^{-7}(1 - 1.6\times10^{-4}) \approx 1.256\times10^{-6} T m/A — just below μ0\mu_0.

Example 32. To get B = 0.5 T in an empty solenoid of 400 turns/m already carrying 1.0 A, what extra (magnetising) current is needed?

Solution: I+Im=Bμ0n=0.54π×10−7×400≈995I + I_m = \frac{B}{\mu_0 n} = \frac{0.5}{4\pi\times10^{-7}\times400} \approx 995 A, so Im≈994I_m \approx 994 A. A suitable high-permeability iron core can greatly reduce the required free current by contributing magnetisation, though real cores may saturate.

Solved Examples - Classifying Materials

Example 33. Classify: χA=+3×10−5\chi_A = +3 \times 10^{-5}, χB=−1\chi_B = -1, χC=+5000\chi_C = +5000, χD=−8×10−6\chi_D = -8\times10^{-6}.

Solution: A: paramagnetic; B: superconductor/perfect diamagnet in the ideal Meissner state (μr=0\mu_r = 0); C: ferromagnetic; D: diamagnetic.

Example 34. A superconducting slab is placed in field intensity H=2×103H = 2\times10^3 A/m. Find M and B inside.

Solution: M=−H=−2×103M = -H = -2\times10^3 A/m; B=μ0(H+M)=0B = \mu_0(H+M) = 0 — Meissner effect, complete expulsion.