Magnetic Effects of Current
1. Introduction
Moving charges produce magnetic fields. This chapter explores the relationship between electricity and magnetism, including field calculations, forces on conductors, and measuring instruments.
2. Biot-Savart Law
dB = (μ₀/4π) × (Idl × r̂)/r²
The magnetic field due to a small current element is proportional to the current, length of element, and sin of angle between element and position vector, and inversely proportional to square of distance.
2.1 Applications
Field at centre of circular loop: B = μ₀I/2R Field on axis of circular loop: B = μ₀IR²/2(R² + x²)^{3/2} Field due to straight conductor: B = μ₀I/4πa (for infinite wire: B = μ₀I/2πa)
3. Ampere's Circuital Law
∮ B · dl = μ₀I_enclosed
Used to find B in symmetric current distributions.
Solenoid: B = μ₀nI (inside an ideal solenoid) Toroid: B = μ₀NI/2πr
4. Force on a Current-Carrying Conductor
F = I(l × B) — the force on a straight conductor. F = q(v × B) — the Lorentz force on a moving charge.
5. Torque on a Current Loop
τ = NIAB sin θ, where θ is the angle between the magnetic field and the normal to the loop.
τ = m × B, where m = NIA is the magnetic moment.
6. Moving Coil Galvanometer
A sensitive instrument for detecting and measuring small currents.
Deflection θ = (NBA/k)I, where k is the torsional constant of the spring.
Conversion to ammeter: Connect a small shunt resistance S = I_gG/(I - I_g) in parallel. Conversion to voltmeter: Connect a large series resistance R = V/I_g - G.
7. Worked Problems
Problem 1: A circular coil of 100 turns and radius 5 cm carries 2 A current. Find B at its centre. Solution: B = μ₀NI/2R = (4π×10^{-7}×100×2)/(2×0.05) = 8π×10^{-4} / 0.1 = 2.51×10^{-3} T.
Problem 2: A solenoid of length 50 cm has 2000 turns and carries 3 A. Find B inside. Solution: n = 2000/0.5 = 4000 turns/m. B = μ₀nI = 4π×10^{-7}×4000×3 = 4π×10^{-7}×12000 = 0.015 T.
Problem 3: A galvanometer of resistance 50 Ω gives full-scale deflection for 5 mA. Convert it to an ammeter reading up to 5 A. Solution: S = I_gG/(I - I_g) = (0.005×50)/(5 - 0.005) = 0.25/4.995 = 0.05 Ω.
8. Common Mistakes
'Students often confuse the direction of the magnetic field. Use the right-hand thumb rule: thumb in direction of current, fingers curl in direction of B.'
'For Ampere's law, always choose an Amperian loop that exploits symmetry and passes through the point where B is to be found.'
9. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Biot-Savart law | 4 | 2 |
| Ampere's circuital law | 4 | 2 |
| Force and torque | 3 | 2 |
| Galvanometer conversion | 4 | 3 |
9. Key Formulae Summary
| Quantity | Formula | Notes |
|---|---|---|
| Biot-Savart law | dB = (μ₀/4π) Idl sinθ/r² | Direction: right-hand rule |
| B at centre of circular loop | B = μ₀I/2R | N turns: B = μ₀NI/2R |
| B on axis of circular loop | B = μ₀IR²/2(R²+x²)^{3/2} | At centre (x=0), matches above |
| B due to infinite wire | B = μ₀I/2πa | a = perpendicular distance |
| B inside solenoid | B = μ₀nI | n = turns per unit length |
| B inside toroid | B = μ₀NI/2πr | N = total turns |
| Force on conductor | F = BIl sinθ | θ = angle between I and B |
| Force between parallel wires | F/l = μ₀I₁I₂/2πd | Attract if same direction |
| Torque on loop | τ = NIAB sinθ | m = NIA (magnetic moment) |
| Galvanometer to ammeter | S = I_gG/(I - I_g) | Shunt in parallel |
| Galvanometer to voltmeter | R = V/I_g - G | Series resistance |
10. Ampere's Law Applications in Detail
10.1 Infinite Straight Conductor
Consider a circular Amperian loop of radius r around the wire. By symmetry, B is tangential and constant.
∮ B·dl = B·2πr = μ₀I ⇒ B = μ₀I/2πr
10.2 Solenoid
For an ideal solenoid of length L with N turns, consider a rectangular Amperian loop. The field is uniform inside and negligible outside.
BL = μ₀NI ⇒ B = μ₀NI/L = μ₀nI
10.3 Toroid
For a toroid of mean radius r, consider a circular Amperian loop inside the toroid:
B·2πr = μ₀NI ⇒ B = μ₀NI/2πr Outside the toroid, net current enclosed is zero, so B = 0.
11. Self-Test Questions
- Derive the expression for B at the centre of a circular current-carrying loop.
- Two long parallel wires carry currents of 5 A and 10 A in the same direction, 10 cm apart. Find force per unit length between them.
- A galvanometer of 100 Ω gives full deflection at 10 mA. Convert it to a voltmeter of range 0-100V.
- Derive the expression for the torque experienced by a current-carrying loop in a uniform magnetic field.
- Using Ampere's circuital law, find the magnetic field due to a toroid.
