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MHT-CET Physics · Magnetic Materials

Magnetic Dipole Moment

A magnet's dipole moment is its pole strength times its length (or magnetisation times volume); in a field B it feels a torque MB sin θ, stores energy −MB cos θ and oscillates with period 2π√(I/MB); an orbiting electron has a moment proportional to its angular momentum, e/2m times L.

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Why this matters

10 PYQs, one HARD. Seven are about magnets: the moment of a rod or a bent rod, a magnet cut into pieces, work to turn a magnet, the period of oscillation, and the torque to hold a coil in a solenoid. Three are the electron's orbital moment — the gyromagnetic ratio and the Bohr magneton. Two cards.

Concept 1 of 2: Moment, Torque, Work and Oscillation

The moment M = m × 2l points from S to N. Bending a rod of length L into a semicircle keeps its pole strength but brings the poles to a diameter 2L/π apart, so the moment falls to 2M/π. Cutting a magnet ALONG its axis halves the pole strength and keeps the length; two such halves placed at right angles give a resultant M/√2. Turning a magnet from the field direction through θ takes work MB(1 − cos θ), so 90° takes twice the work of 60°. It oscillates with T = 2π√(I/MB): stronger field or moment, shorter period. Two magnets held together add their moments with like poles together and subtract them with unlike poles together. A current loop is a dipole too: τ = NIAB for a coil whose axis is perpendicular to B.

Definition

  • M=m×2l=M = m \times 2l = magnetisation × volume (5 cm, 1 cm diameter, 5.3×1035.3\times10^3 A/m ⇒ 2×10−22\times10^{-2} J/T).
  • Bent into a semicircle: 2Mπ\dfrac{2M}{\pi}. Cut along the axis, halves at 90°: M2\dfrac{M}{\sqrt{2}}.
  • Work from the field direction: W=MB(1−cos⁡θ)W = MB(1 - \cos\theta) (90° : 60° = 2 : 1).
  • Oscillation: T=2πIMBT = 2\pi\sqrt{\dfrac{I}{MB}}; field tripled ⇒ T÷3T \div \sqrt{3}. Magnets 2M2M and MM held together: like poles together (moments add) 3M3M, unlike poles together MM ⇒ T1:T2=1:3T_1 : T_2 = 1 : \sqrt{3}.
  • Coil in a solenoid: τ=NIAB\tau = NIAB, B=μ0nIB = \mu_0 nI.

Magnet in a field

τ=MBsin⁡θ,W=MB(1−cos⁡θ),T=2πIMB\tau = MB\sin\theta, \qquad W = MB(1 - \cos\theta), \qquad T = 2\pi\sqrt{\frac{I}{MB}}

Worked example

A magnet of moment 2 A m² lies along a 0.4 T field. Work to turn it through 90°, and the torque in that position?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q32Moderate

Example 1 · Magnetic Materials · Magnetic Dipole Moment and Bar Magnet

Two bar magnets A and B are geometrically similar, but the magnetic moment of A is twice that of B.T1B.T_{1} is the time period of oscillation when their like poles are kept together. When unlike poles are kept together, the time period of oscillation is T2T_{2}. The ratio T1:T2T_{1}:T_{2} will be

Keeping the length when a rod is bent

The moment uses the straight distance between the poles. A rod bent into a semicircle has its poles a diameter 2L/π apart, not L.

Subtracting moments of magnets held with like poles together

Like poles together, the magnets point the same way and the moments ADD: 2M and M give 3M, so the period is SHORTER. Unlike poles together, they subtract to M.
Drill 6 more on moment, torque, work and oscillation

Concept 2 of 2: The Orbiting Electron's Moment

An electron circling at frequency f is a current ef around an area πr², so its moment is (e/2m)L: proportional to its angular momentum, pointing opposite to it because the electron is negative. The ratio e/2m is the gyromagnetic ratio; with L = h/2π in the first Bohr orbit the moment is the Bohr magneton eh/4πm.

Definition

  • μ=e2mL\mu = \dfrac{e}{2m}L, directed opposite to L: μ⃗=−RL⃗\vec{\mu} = -R\vec{L} with R=e2mR = \dfrac{e}{2m}.
  • Bohr magneton: μB=eh4πm≈9.27×10−24\mu_B = \dfrac{eh}{4\pi m} \approx 9.27\times10^{-24} A m².

Orbital moment

μ=e2mL,μB=eh4πm\mu = \frac{e}{2m}L, \qquad \mu_B = \frac{eh}{4\pi m}

Worked example

An electron's orbital angular momentum is 2h/2π. Its orbital magnetic moment in Bohr magnetons?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q1Moderate

Example 2 · Magnetic Materials · Magnetic Dipole Moment and Bar Magnet

The gyromagnetic ratio and Bohr magneton are given respectively by [Given: ee = charge on electron, mm = mass of electron, hh = Planck's constant]

Writing the gyromagnetic ratio as e/m

The loop's area and current give μ/L = e/2m — half of e/m.

Dropping the minus sign

The electron is negative, so its orbital moment points OPPOSITE its angular momentum: μ = −(e/2m)L.
Drill 2 more on the orbiting electron's moment

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Moment, Torque, Work and Oscillation

    Magnet in a field

    τ=MBsin⁡θ,W=MB(1−cos⁡θ),T=2πIMB\tau = MB\sin\theta, \qquad W = MB(1 - \cos\theta), \qquad T = 2\pi\sqrt{\frac{I}{MB}}
  • The Orbiting Electron's Moment

    Orbital moment

    μ=e2mL,μB=eh4πm\mu = \frac{e}{2m}L, \qquad \mu_B = \frac{eh}{4\pi m}

Watch out for (4)

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