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MHT-CET Physics · Magnetic Fields Due to Electric Current

Magnetic Moment, Torque on a Loop, and the Galvanometer

A current loop behaves like a small magnet of moment m = NIA, feeling a torque m × B in a field; for a fixed length of wire the torque is largest when the loop is a circle, an orbiting electron is a tiny loop whose moment is e/(2m) times its angular momentum, and a galvanometer becomes an ammeter when a small shunt takes most of the current.

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Why this matters

20 PYQs, 5 HARD. Seventeen are the magnetic moment — its value from the field at the centre and the area (the HARD ones), how it changes when a coil is rewound, the torque on a square and a circle made from the same wire, a solenoid's moment from its flux, and an orbiting electron's moment; three are shunts — the shunt for a given fraction of the current, and an ammeter against a milliammeter. Two cards.

Concept 1 of 2: Magnetic Moment and the Torque on a Loop

m = NIA, and in a field the loop feels τ = mB sin θ. For the same length of wire, a circle encloses more area than a square, so it feels more torque; bent into one circular turn of wire length L it has area L²/(4π), so τ_max = L²IB/(4π). Rewinding a coil into a smaller radius R/3 triples the turns but cuts each area to a ninth: the moment falls to a third. Combine with the field at the centre, B = μ₀I/(2R): then m = 2πBR³/μ₀ = 2BA^(3/2)/(μ₀√π), and the ratio B/m = μ₀/(2πR³) falls eightfold when R doubles. An orbiting electron's moment is (e/2m) times its angular momentum, so it grows with n.

Definition

  • m=NIAm = NIA (depends on N, I and r); τ=mBsin⁡θ\tau = mB\sin\theta (or r⃗×F⃗\vec r \times \vec F for a single force: (4i^−3j^)×(5i^−10j^)=−25k^(4\hat i - 3\hat j) \times (5\hat i - 10\hat j) = -25\hat k).
  • Same wire, square vs circle: the circle has the larger area and torque. One turn of wire length L: τmax⁡=L2IB4π\tau_{\max} = \dfrac{L^2IB}{4\pi}; m=IL24πm = \dfrac{IL^2}{4\pi} ⇒ L=4mπIL = 4\sqrt{\dfrac{m}{\pi I}}.
  • Rewound to R/3: m′=m3m' = \tfrac{m}{3}. Equal moments with radii 10 cm and 20 cm: NAIA=4NBIBN_AI_A = 4N_BI_B.
  • From the field at the centre: m=2πBR3μ0=2BA3/2μ0πm = \dfrac{2\pi BR^3}{\mu_0} = \dfrac{2BA^{3/2}}{\mu_0\sqrt{\pi}}; Bm=μ02πR3\dfrac{B}{m} = \dfrac{\mu_0}{2\pi R^3}.
  • Solenoid: NIA=ϕlμ0NIA = \dfrac{\phi l}{\mu_0} (1.57 × 10⁻⁶ Wb, 0.8 m ⇒ 1 A m²).
  • Orbiting electron: Lmorb=2me\dfrac{L}{m_{\text{orb}}} = \dfrac{2m}{e}; morb=neh4πm∝nm_{\text{orb}} = \dfrac{neh}{4\pi m} \propto n.

Magnetic moment

m=NIA,τ=mBsin⁡θ,m=2πBR3μ0m = NIA, \qquad \tau = mB\sin\theta, \qquad m = \frac{2\pi BR^3}{\mu_0}

Worked example

A 50-turn coil of radius 4 cm carries 2 A in a 0.3 T field, its plane parallel to the field. Moment and torque?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q49Hard

Example 1 · Magnetic Fields Due to Electric Current · Magnetic Moment of Current Loop and Galvanometer Instruments

Magnetic field at the centre of a circular loop of area 'A' is 'B'. The magnetic moment of the loop will be

Keeping the turns fixed when a coil is rewound

The wire's length is fixed: a coil rewound to a third of the radius has three times the turns. m = NIπr² then falls to a third, not a ninth.
Drill 16 more on magnetic moment and the torque on a loop

Concept 2 of 2: Converting a Galvanometer: the Shunt

A galvanometer carries only a small current I_g at full deflection. To measure a larger current I, a low resistance S in parallel (a shunt) takes the rest: both share the same voltage, so I_gG = (I − I_g)S and S = G/(I/I_g − 1). The larger the range, the SMALLER the shunt — an ammeter's shunt is less than a milliammeter's.

Definition

  • S=IgGI−Ig=GIIg−1S = \dfrac{I_gG}{I - I_g} = \dfrac{G}{\tfrac{I}{I_g} - 1}.
  • 4% through the galvanometer ⇒ S=G24S = \dfrac{G}{24}; 3% through 200 Ω ⇒ S=0.03×2000.97≈6S = \dfrac{0.03 \times 200}{0.97} \approx 6 Ω.
  • Higher range ⇒ smaller shunt.

Shunt

S=GIIg−1S = \frac{G}{\frac{I}{I_g} - 1}

Worked example

A 50 Ω galvanometer reads full scale at 2 mA. Shunt to make it a 1 A ammeter?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q41Moderate

Example 2 · Magnetic Fields Due to Electric Current · Magnetic Moment of Current Loop and Galvanometer Instruments

A galvanometer of resistance 200 Ω200\,\Omega is to be converted into an ammeter. The value of shunt resistance which allows 3% of the mains current through the galvanometer is equal to (nearly)

Using the total current over the galvanometer current

S = G/(I/I_g − 1): subtract 1, because the galvanometer still carries its own share. 4% gives G/24, not G/25.
Drill 2 more on converting a galvanometer: the shunt

Summary — formulas & gotchas at a glance

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Formulas (2)

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