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PhysicsMagnetic Effects of Current
A 2 amp current is flowing through two different small circular copper coils having radii ratio 1:2. The ratio of their respective magnetic moments will be:
Options
1
4:1
2
1:4
3
1:2
4
2:1
Correct Answer
1:4
Solution
1

Magnetic moment: M = IA = I x pi r^2

Same current I in both coils.

M proportional to r^2

2

Radii ratio r1:r2 = 1:2

M1/M2 = r1^2/r2^2 = (1)^2/(2)^2 = 1/4

Ratio M1:M2 = 1:4

M = IA = I x pi x r^2 | M proportional to r^2
M1:M2 = 1^2 : 2^2 = 1:4
Theory: Magnetic Effects of Current
1. Magnetic Moment of Current Loop

A current-carrying loop acts as a magnetic dipole. Magnetic moment M = NIA. N = number of turns, I = current (A), A = area (m^2). For circular coil: A = pi r^2. Direction: perpendicular to plane, right-hand rule (curl fingers in current direction, thumb points along M). M is a vector quantity. In external magnetic field B: torque tau = M x B. Potential energy U = -M.B.

2. Magnetic Field at Centre of Circular Loop

B at centre = mu0 I / (2R). For N turns: B = mu0 N I / (2R). Inversely proportional to radius. At axis at distance x from centre: B = mu0 I R^2 / (2(R^2 + x^2)^(3/2)). At large distance (x >> R): B = mu0 M / (2pi x^3) - looks like a dipole field.

3. Ampere Circuital Law

Integral of B.dl around any closed loop = mu0 I_enclosed. Useful for symmetric situations. For a long straight wire: B = mu0 I / (2pi r). For solenoid: B = mu0 n I (n = turns per unit length). For toroid: B = mu0 N I / (2pi r) inside, B = 0 outside. Ampere law is the magnetostatic equivalent of Gauss law.

4. Biot-Savart Law

dB = (mu0/4pi)(I dl x r_hat)/r^2. Used when Ampere law cannot be applied (non-symmetric configurations). For circular loop at centre: B = mu0 I/(2R). For infinite straight wire: B = mu0 I/(2pi r) (same as Ampere law, consistent). For finite wire: B = (mu0 I/4pi r)(sin theta1 + sin theta2).

5. Magnetic Dipole in External Field

Torque: tau = MB sin theta = M x B. For theta = 90 deg: maximum torque tau = MB. For theta = 0 or 180 deg: zero torque (stable/unstable equilibrium). Work done rotating from theta1 to theta2: W = MB(cos theta1 - cos theta2). Potential energy: U = -MB cos theta = -M.B. Analogy: exactly like electric dipole p in electric field E, replacing p with M and E with B.

6. Galvanometer

Moving coil galvanometer: current-carrying coil in radial magnetic field. Torque due to B: tau = NIAB (radial field ensures constant torque). Restoring torque from spiral spring: tau = C theta. Equilibrium: NIAB = C theta. Deflection theta = NIAB/C. Sensitivity: theta/I = NAB/C. To convert to ammeter: add low resistance shunt in parallel. To convert to voltmeter: add high resistance in series. Ballistic galvanometer: measures charge (first-throw deflection proportional to charge).

7. Bohr Magneton

Smallest unit of magnetic moment for electron. mu_B = eh/(4pi m_e) = 9.27 x 10^-24 J/T. Orbital magnetic moment of electron in nth Bohr orbit: M = n x mu_B. Spin magnetic moment of electron: M = sqrt(s(s+1)) x g_s x mu_B ≈ mu_B (g_s = 2 for electron). Total magnetic moment determines paramagnetism, ferromagnetism of materials.

8. Magnetic Properties of Materials

Diamagnetic: weak repulsion from B field. All materials have diamagnetism. Induced moments oppose field. Examples: Bi, Cu, water. Paramagnetic: weak attraction. Random magnetic dipoles partially align with B field. Examples: Al, Mn, O2. Ferromagnetic: strong attraction. Magnetic domains align spontaneously. Hysteresis loop. Examples: Fe, Ni, Co. Retentivity and coercivity important for permanent magnets. Curie temperature: above this ferromagnets become paramagnetic.

Frequently Asked Questions
1. Why does M scale with r^2 and not r?
Magnetic moment M = IA = I x pi x r^2. The area A of a circle scales as r^2 (area = pi r^2). Doubling the radius quadruples the area and thus quadruples the magnetic moment (for the same current). This is a geometrical consequence. Unlike the magnetic field at the centre (which scales as 1/r), the magnetic moment involves the area, not the circumference.
2. What is the magnetic field at the centre of each coil?
B = mu0 I / (2R). For coil 1: B1 = mu0 I / (2R). For coil 2: B2 = mu0 I / (2 x 2R) = mu0 I / (4R). Ratio B1/B2 = 2:1. The smaller coil has stronger field at centre. Contrast: magnetic moment ratio 1:4 (larger coil has larger moment). This shows that M and B at centre are inversely related for same current.
3. What is the difference between magnetic moment and magnetic field?
Magnetic moment M = NIA: a property of the current loop itself (how strong a magnetic source it is). Units: A m^2. Magnetic field B = mu0 I/(2R) at centre: the field produced at a specific point. Units: Tesla. They scale differently with radius: M proportional to r^2, B at centre proportional to 1/r. A large coil has large M but small B at centre; a small coil has small M but large B at centre.
4. How is magnetic moment used in MRI?
MRI (Magnetic Resonance Imaging) uses the magnetic moment of hydrogen nuclei (protons). Protons have intrinsic spin magnetic moment. In a strong external B field (1.5-3 T), proton magnetic moments align. Radio frequency pulses knock them out of alignment. As they realign, they emit RF signals. Different tissues have different proton density and relaxation times, giving contrast. The fundamental physics is the interaction of magnetic moment with external B field: E = -M.B.
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