Question

Given \( n \) turns, current \( I \), permeability, and diameter, find magnetisation.

Hint:

Magnetisation is calculated as the number of turns per unit length times the current in the coil.

Answer 1

Step 1: Understanding magnetisation.
Magnetisation \( M \) is defined as the magnetic moment per unit volume of a material. For a coil with \( n \) turns carrying a current \( I \), the magnetisation is given by the product of the number of turns per unit length and the current: \[ M = nI \] Step 2: Explanation.
In this formula, \( n \) is the number of turns per unit length (i.e., the number of loops per unit length of the coil), and \( I \) is the current passing through the coil. The product \( nI \) gives the magnetisation of the material. This formula is valid for a solenoid or a coil where the turns are uniformly distributed.
Step 3: Conclusion.
Thus, the magnetisation is given by \( M = nI \), where \( n \) is the number of turns per unit length, and \( I \) is the current.

Answer 2

Step 1: Magnetisation is the magnetic moment produced per unit volume of a material.

Step 2: For a coil with n turns per unit length carrying current I, the magnetisation is simply M = n × I.

Step 3: So just multiply the number of turns per unit length by the current flowing through the coil.

Answer 3

Via the solenoid field relation.
Inside a long current-carrying solenoid, the magnetic field is \( B = \mu_0 n I \), where \( n \) is the number of turns per unit length and \( I \) is the current.
Magnetisation is related to the field it produces by \( M = B/\mu_0 \) (for the field generated purely by the winding current, ignoring the core's own response).
Substituting, \[ M = \frac{\mu_0 n I}{\mu_0} = nI \] which agrees with the direct definition of magnetisation as magnetic moment per unit volume.