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Solenoid

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Estimated time: 13 minutes
CBSE: Class 12

Definition: Solenoid

A long solenoid is a coil whose length is much greater than its radius, producing a uniform magnetic field inside and nearly zero field outside.

OR

A solenoid is a long helical coil of insulated wire with many closely spaced turns, whose length l is much greater than its radius R (i.e., l ≫ R), such that it produces a strong, uniform magnetic field inside and a negligible field outside.

CBSE: Class 12

Formula: Magnetic Field Inside a Long Solenoid

B = μ0nI

Where:

  • μ0 = permeability of free space
  • n = number of turns per unit length
  • I = current
CBSE: Class 12

Construction

  • A solenoid consists of a long cylindrical former (can be air-core, iron-core, or insulating rod).
  • Enamelled copper wire is wound tightly in helical turns — insulation prevents short-circuiting between adjacent turns.
  • Total turns: N; Length: l; Turns per unit length: n = \[\frac  {N}{l}\]
  • When current I flows → magnetic field is produced inside the windings.
CBSE: Class 12

Working Principle

  • Each turn of the solenoid acts as a small current loop, producing a small magnetic field along the axis.
  • In a long solenoid, the individual fields of each turn superpose constructively along the axis → uniform field inside.
  • Outside the solenoid, fields from adjacent turns are in opposite directions and cancel → field ≈ 0 outside.
  • The net effect: the solenoid behaves like a bar magnet with polarity determined by the direction of current (Right-Hand Thumb Rule).
CBSE: Class 12

Derivation of Magnetic Field Using Ampere's Law

Step 1 -S Set up the Amperian Loop:
Consider a rectangular Amperian loop abcd of length l, with side ab inside the solenoid (along the axis) and side cd outside.

Step 2 - Evaluate \[\oint\vec{B}\cdot d\vec{l}\] for each side:

Side Condition Contribution
ab (inside, along axis) \[\vec B\] || \[d\vec l\] Bl
bc (perpendicular, inside→outside) \[\vec B\] ⊥ \[d\vec l\] 0
cd (outside solenoid) Boutside ≈ 0 0
da (perpendicular, outside→inside) \[\vec B\] ⊥ \[d\vec l\] 0

\[\therefore\oint\vec{B}\cdot d\vec{l}\] = Bl

Step 3 - Calculate the enclosed current:

Number of turns inside loop = nl
∴ Total enclosed current = Ienc = n ⋅ l ⋅ I

Step 4 - Apply Ampere's Law:

Bl = μ0 ⋅ nlI
B = μ0nI
CBSE: Class 12

Special Cases

Location Magnetic Field Reason
Inside (ideal/infinite solenoid) B = μ0nI Superposition of all turns
At the ends (edge) B = \[\frac {μ_0nI}{2}\] Only half the turns contribute
Outside B ≈ 0 Cancellation of adjacent fields
With an iron core B = μ0μrnI= Core amplifies permeability
CBSE: Class 12

Solenoid vs. Toroid

Feature Solenoid Toroid
Shape Long straight cylinder Solenoid bent into a closed ring (doughnut)
Field inside B = μ0nI (uniform) B = \[\frac {μ_0NI}{2πr}\] (varies with r)
Field outside B ≈ 0 (not exactly zero) B = 0 (exactly zero)
Field at ends B = \[\frac {μ_0nI}{2}\] No ends (closed loop)
Uniformity Uniform inside (away from ends) Not uniform; depends on radius
Application Electromagnets, relays, MRI Inductors, transformers, RF circuits
CBSE: Class 12

Applications

  • MRI Machines — superconducting solenoids produce the powerful, stable magnetic field required for imaging.
  • Electric Motors and Generators — solenoids create the rotating magnetic fields needed for motor operation.
  • Solenoid Valves — used in hydraulic and pneumatic systems to control fluid flow (e.g., washing machines, car engines).
  • Circuit Breakers / Relays — electromagnetic solenoids open/close switches on detecting fault current.
  • Hard Disks and Loudspeakers — solenoid-based actuators position read/write heads and drive speaker cones.
CBSE: Class 12

Example

Given:

  • Length of solenoid: l = 0.5 m

  • Radius: r = 1 cm = 0.01 m

  • Total number of turns: N = 500

  • Current: I = 5 A

Step 1- Find turns per unit length (n):

n = \[\frac {N}{l}\] = \[\frac {500}{0.5}\] = 1000 turns/m

Step 2 - Check if it qualifies as a long solenoid:

\[\frac {l}{r}\] = \[\frac {0.5}{0.01}\] = 50

Since l/r = 50 ≫ 1, i.e., l ≫ r, the solenoid qualifies as a long solenoid, and we can use the standard formula.

Step 3 - Apply the formula B = μ0nI:

B = (4π × 10−7) × 1000 × 5
B = 4π × 10−7 × 5000 = 6.28 × 10−3 T

Answer: The magnetic field inside the solenoid is approximately 6.28 × 10−3 T (or ≈ 6.3 × 10−3 T).

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