To demonstrate the similarity of a current carrying solenoid to a bar magnet, let us calculate axial field of a finite solenoid carrying current.

Consider: ** a** = radius of solenoid

** 2l **= length of solenoid with centre O

** n** = number of turns per unit length

** I **= current passing through solenoid

*OP = r *

Consider a small element of thickness ** dx** of solenoid at distance

**from O. and number of turns in element**

*x*

*= n dx.*We know magnetic field due to n turns coil at axis of solenoid is given by

The magnitude of the total field is obtained by summing over all the elements — in other words by integrating from x = – l to x = + l. Thus,

This integration can be done by trigonometric substitutions. This exercise, however, is not necessary for our purpose. Note that the range of x is from – l to + l. Consider the far axial field of the solenoid, i.e., r >> a and r >> l. Then the denominator is approximated by

Note that the magnitude of the magnetic moment of the solenoid is, m = n (2l) I (a2) — (total number of turns × current × cross-sectional area). Thus,

It is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

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