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What is the magnetic field due to a solenoid on its axis?

Published in Electromagnetism 3 mins read

The magnetic field due to a solenoid on its axis is directly proportional to the permeability of the material, the number of turns per unit length, and the current passing through the solenoid.

The magnetic field (B) inside a solenoid is relatively uniform and its direction is along the axis of the solenoid, determined by the right-hand rule. The formula for the magnetic field (B) on the axis of an ideal solenoid is given by:

B = μnI

Where:

  • B is the magnetic field strength (in Tesla).
  • μ is the permeability of the core material (in Tesla meters per Ampere). If the core is air or a vacuum, this is μ0 (the permeability of free space ≈ 4π × 10-7 T⋅m/A).
  • n is the number of turns of wire per unit length (turns per meter).
  • I is the current flowing through the wire (in Amperes).

Understanding the Components

Let's break down each component of the formula to understand its impact on the magnetic field:

  • Permeability (μ): The higher the permeability of the core material, the stronger the magnetic field. Ferromagnetic materials like iron have much higher permeability than air, thus significantly increasing the magnetic field strength.

  • Turns per unit length (n): This represents the density of the coil windings. A solenoid with more turns packed tightly together will produce a stronger magnetic field than one with fewer, more loosely spaced turns. For example, a solenoid that is 1 meter long and has 1000 turns has n = 1000 turns/meter.

  • Current (I): The magnetic field strength is directly proportional to the current flowing through the solenoid wire. Doubling the current doubles the magnetic field, assuming all other factors remain constant.

Direction of the Magnetic Field

The direction of the magnetic field is along the axis of the solenoid. The right-hand rule can be used to determine the direction:

  1. Right-Hand Rule: If you curl the fingers of your right hand in the direction of the current flow in the solenoid coils, your thumb will point in the direction of the magnetic field inside the solenoid.

Example Calculation

Let's consider a solenoid with the following characteristics:

  • Number of turns: 500 turns
  • Length: 0.25 meters
  • Current: 2 Amperes
  • Core Material: Air (μ = μ0 ≈ 4π × 10-7 T⋅m/A)

First, calculate n:

  • n = Number of turns / Length = 500 turns / 0.25 m = 2000 turns/m

Now, use the formula to find the magnetic field:

  • B = (4π × 10-7 T⋅m/A) (2000 turns/m) (2 A)
  • B ≈ 5.03 × 10-3 Tesla

Therefore, the magnetic field on the axis of this solenoid is approximately 5.03 × 10-3 Tesla.

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