Applications of Ampere’s Circuital Law > Magnetic Field of a Long Straight Solenoid

• The magnetic field pattern of a solenoid closely resembles that of a bar magnet.
• One end behaves as the North pole, the other as the South pole.
• Inside the solenoid, magnetic field lines are parallel, indicating a uniform magnetic field.

A long, cylindrical coil consisting of a large number of closely wound circular turns of insulated copper wire, in which a magnetic field is produced when current flows through it.

Given: An ideal long solenoid with n turns per unit length, carrying current I.

Step-by-step derivation:

1. Consider a rectangular Amperian loop abcd, with side ab (length l) inside the solenoid, parallel to the axis.
2. Apply Ampere's Circuital Law: \[\oint\vec{B}\cdot d\vec{l}=\mu_0I_{enc}\]
3. The field outside a long solenoid is negligible, and the contributions from sides bc and da (perpendicular to B) are zero.
4. The line integral reduces to: B ⋅ l = μ₀I_enc
5. Total enclosed current: I_enc = nlI (since there are nl turns enclosed).
6. Substituting: Bl = μ₀(nlI)

Final Result: B = μ₀nI

For a solenoid with a magnetic core (relative permeability μ_r​): B = μ₀μ_rnI

Given:

• Length of solenoid, L = 50 cm = 0.5 m
• Mean radius, r = 1.4 cm
• Number of layers = 4, turns per layer = 350 → Total turns N = 1400
• Current, I = 6.0 A

Find: Magnetic field at the centre and at the end of the solenoid.

Solution:

Step 1: Turns per unit length: n = \[\frac {N}{L}\] = \[\frac {1400}{0.5}\] = 2800 m⁻¹

Step 2: Field at centre: B = μ₀nI = (4π × 10⁻⁷)(2800)(6.0) = 2.1 × 10⁻² T

Step 3: Field at the end: B_end = \[\frac {1}{2}\]μ₀nI = 1.05 × 10⁻² T

Final Answer: B_center = 2.1 × 10⁻² T, B_end = 1.05 × 10⁻² T

• Analogy: A solenoid behaves like a bar magnet — comparing field-line patterns side by side helps visualize pole formation at the two ends.
• Applications: Electromagnets (used in relays, circuit breakers), MRI machine coils, loudspeakers, and solenoid valves in automotive/industrial systems.