What is the magnetic field inside a solenoid?

The magnetic field inside a long solenoid is uniform and given by B = μ₀nI, where n is the number of turns per unit length (n = N/L) and I is the current. Outside an ideal solenoid, the field is zero.

How does a moving coil galvanometer work?

A moving coil galvanometer works on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The deflecting torque τ = BINA balances the restoring torque of the spring τ = kθ. At equilibrium: θ = (BINA/k) × I, so deflection is proportional to current.

How do you convert a galvanometer to an ammeter?

A galvanometer is converted to an ammeter by connecting a low-resistance shunt S in parallel. The shunt value: S = Ig × G / (I − Ig), where Ig is the full-scale deflection current of the galvanometer, G is galvanometer resistance, and I is the maximum current to be measured.

What is Ampere's Circuital Law?

Ampere's Circuital Law states that the line integral of the magnetic field B around any closed loop (Amperian loop) equals μ₀ times the net current passing through the surface bounded by the loop: ∮B·dl = μ₀I_enclosed.

Tamil Nadu Class 12 Physics — Magnetism and Magnetic Effects Notes, Formulae & Important Questions

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

Board Exam Weight: Magnetism and Magnetic Effects carries 8–10 marks in Tamil Nadu Class 12 Physics board exams (Volume 1, Chapter 3). NEET typically has 3–4 questions from this chapter.

1. Magnetic Field — Concept

A magnetic field B exists in the region around a magnet or a current-carrying conductor. Its SI unit is the Tesla (T). It is a vector quantity.

Lorentz Force on a moving charge in a magnetic field:

F = qv × B   →   |F| = qvB sinθ

Maximum when v ⊥ B (θ = 90°): F = qvB
Zero when v ∥ B (θ = 0°)
Force is always perpendicular to both v and B

2. Biot-Savart Law

The magnetic field due to a small current element Idl at a distance r:

dB = (μ₀/4π) × (Idl sinθ) / r²

where:

μ₀ = 4π×10⁻⁷ T·m/A (permeability of free space)
θ = angle between current direction and position vector r
Direction: given by right-hand rule (curl fingers from dl toward r̂, thumb gives B direction)

3. Magnetic Field Due to Common Configurations

Long Straight Wire (distance r from wire)

Using Biot-Savart (or Ampere's law):

B = μ₀I / (2πr)

Direction: Concentric circles around wire — right-hand rule (thumb along current, fingers curl in B direction).

Circular Coil (field at centre)

For N turns of radius R:

B = μ₀NI / (2R)

Direction: Perpendicular to the plane of the coil (right-hand rule for the coil).

Solenoid (field inside)

For a solenoid with n turns per unit length (n = N/L):

B = μ₀nI

The field is uniform inside and zero outside an ideal solenoid.

Toroid

B = μ₀NI / (2πr)    [inside the toroid]
B = 0                 [outside]

4. Ampere's Circuital Law

"The line integral of B around any closed loop equals μ₀ times the net current enclosed":

∮ B · dl = μ₀ I_enclosed

Applications:

Derives field of long straight wire: B = μ₀I/(2πr)
Derives field inside solenoid: B = μ₀nI
Derives field inside toroid

5. Force on a Current-Carrying Conductor

A conductor of length L carrying current I in a magnetic field B:

F = BIL sinθ

where θ is the angle between the current direction and B. Maximum force when θ = 90°.

Direction: Fleming's left-hand rule — First finger: B, Second finger: I, Thumb: Force direction.

Force Between Two Parallel Wires

Two parallel wires carrying currents I₁ and I₂, separated by distance d:

F/L = μ₀ I₁ I₂ / (2πd)   (force per unit length)

Same direction currents: Attractive
Opposite direction currents: Repulsive

6. Torque on a Current Loop

A rectangular coil of N turns, area A, carrying current I, placed in field B at angle α between the plane of coil and B:

τ = BINA sinα

When α = 90° (coil perpendicular to B), torque is maximum: τ_max = BINA. When α = 0° (coil parallel to B), torque is zero.

7. Moving Coil Galvanometer

Principle: A current-carrying coil in a magnetic field experiences a deflecting torque.

Construction: Coil of N turns on a soft-iron core, suspended between permanent magnet poles, with a spiral spring providing restoring torque.

Working

Deflecting torque: τ_d = BINA (radial field ensures constant torque for all positions)
Restoring torque: τ_r = kθ (k = spring constant)
At equilibrium: BINA = kθ

θ = (BNA/k) × I

Deflection θ is directly proportional to current I.

Current Sensitivity: S_I = θ/I = BNA/k (rad/A)

Voltage Sensitivity: S_V = θ/V = BNA/(kG) where G = coil resistance

Conversion to Ammeter

Connect a low-resistance shunt S in parallel:

S = (Ig × G) / (I − Ig)

where Ig = full-scale deflection current, G = galvanometer resistance, I = range of ammeter.

An ammeter has very low resistance and is connected in series.

Conversion to Voltmeter

Connect a high-resistance multiplier R_m in series:

R_m = (V/Ig) − G

A voltmeter has very high resistance and is connected in parallel.

8. Worked Numericals

Numerical 1 — Field Due to Long Straight Wire

Problem: A long straight wire carries a current of 5 A. Find the magnetic field at a distance of 10 cm from the wire.

Solution

Given: I = 5 A, r = 10 cm = 0.10 m
μ₀ = 4π×10⁻⁷ T·m/A

B = μ₀I / (2πr)
B = (4π×10⁻⁷ × 5) / (2π × 0.10)
B = (20π×10⁻⁷) / (0.2π)
B = (20×10⁻⁷) / 0.2
B = 100×10⁻⁷ = 1×10⁻⁵ T = 10 μT

Numerical 2 — Circular Coil

Problem: A circular coil of 20 turns, radius 0.1 m, carries a current of 2 A. Find the magnetic field at the centre of the coil.

Solution

Given: N = 20, R = 0.1 m, I = 2 A
μ₀ = 4π×10⁻⁷ T·m/A

B = μ₀NI / (2R)
B = (4π×10⁻⁷ × 20 × 2) / (2 × 0.1)
B = (160π×10⁻⁷) / 0.2
B = 800π×10⁻⁷
B ≈ 2.51×10⁻⁴ T = 0.251 mT

Numerical 3 — Solenoid

Problem: A solenoid has 500 turns per metre and carries a current of 2 A. Find the magnetic field inside the solenoid.

Solution

Given: n = 500 turns/m, I = 2 A
μ₀ = 4π×10⁻⁷ T·m/A

B = μ₀nI
B = 4π×10⁻⁷ × 500 × 2
B = 4π×10⁻⁷ × 1000
B = 4000π×10⁻⁷
B ≈ 1.26×10⁻³ T = 1.26 mT

Numerical 4 — Force on Current-Carrying Conductor

Problem: A wire 2 m long carries a current of 3 A and is placed in a uniform magnetic field of 0.5 T. The wire makes an angle of 30° with the field. Find the force on the wire.

Solution

Given: L = 2 m, I = 3 A, B = 0.5 T, θ = 30°

F = BIL sinθ
F = 0.5 × 3 × 2 × sin30°
F = 3 × 0.5
F = 1.5 N

The force on the wire is 1.5 N, directed perpendicular to both the wire and the field.

9. Board Exam Model Questions

2-Mark Questions

Q1. State Biot-Savart law and give its vector form. Biot-Savart law: The magnetic field dB due to a current element Idl at distance r making angle θ is dB = (μ₀/4π)(Idl sinθ)/r². Vector form: dB = (μ₀/4π) × (I dl × r̂)/r², direction by right-hand rule.

Q2. Distinguish between a galvanometer, an ammeter, and a voltmeter in terms of resistance and connection.

Answer

Instrument	Resistance	Connection
Galvanometer	Moderate (G)	—
Ammeter	Very low	Series
Voltmeter	Very high	Parallel

Q3. What is meant by current sensitivity of a galvanometer? How can it be increased? Current sensitivity = deflection per unit current = BNA/k (rad/A or div/μA). It can be increased by: increasing B (stronger magnets), increasing N (more turns), increasing A (larger coil), or decreasing k (weaker spring).

5-Mark Questions

Q4. Derive the expression for the magnetic field at the centre of a circular current-carrying coil using Biot-Savart law.

Answer

For a small element dl of a circular coil of radius R carrying current I, θ = 90° always (dl ⊥ r̂).

dB = (μ₀/4π) × Idl / R²

All elements contribute in the same direction (perpendicular to plane, by right-hand rule). Integrating over the full circle (total length = 2πR):

B = (μ₀/4π) × I × 2πR / R² = μ₀I / (2R)

For N turns: B = μ₀NI/(2R).

Q5. Explain the construction and working of a moving coil galvanometer. How is it converted to an ammeter?

Answer

Construction: Rectangular coil of N turns wound on a soft-iron core, placed between curved permanent magnet poles (for radial field). Two springs provide restoring torque.

WorkingCurrent I → deflecting torque τ_d = BINA. Spring → restoring torque τ_r = kθ. At equilibrium: θ = (BNA/k)I.

Conversion to ammeter: Connect shunt S = IgG/(I − Ig) in parallel. Most current (I − Ig) bypasses through shunt; only Ig passes through galvanometer.

10. NEET Relevance

Topic	NEET Frequency
Force on moving charge in B field	Very High
Biot-Savart — circular loop	High
Ampere's law — solenoid/long wire	High
Force on current-carrying conductor	High
Torque on current loop	Medium
Galvanometer conversion	Medium

NEET Tip: Know the direction rules thoroughly — right-hand rule, Fleming's left-hand rule, and the screw rule. NEET frequently tests direction-based questions in magnetism.

11. Key Formulae Summary

Quantity	Formula
Lorentz force	F = qvB sinθ
Biot-Savart element	dB = (μ₀/4π) Idl sinθ/r²
Long straight wire	B = μ₀I/(2πr)
Circular coil (centre)	B = μ₀NI/(2R)
Solenoid	B = μ₀nI
Force on conductor	F = BIL sinθ
Torque on coil	τ = BINA sinα
Galvanometer equation	θ = (BNA/k)I
Shunt (ammeter)	S = IgG/(I−Ig)
Multiplier (voltmeter)	R_m = V/Ig − G

Constants: μ₀ = 4π×10⁻⁷ T·m/A, e = 1.6×10⁻¹⁹ C

Frequently Asked Questions

Biot-Savart law states that the magnetic field dB at a point P due to a small current element Idl is: dB = (μ₀/4π) × Idl sinθ / r², where θ is the angle between the current element and the line joining it to point P, and r is the distance. Direction is given by the right-hand rule.