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 x from O. and number of turns in element = 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.

How useful was this post?

Click on a star to rate it!

Average rating 3.5 / 5. Vote count: 110

No votes so far! Be the first to rate this post.

As you found this post useful...

Follow us on social media!

We are sorry that this post was not useful for you!

Let us improve this post!

Tell us how we can improve this post? Please mention your Email so that we can contact you for better feedback.

Leave a Reply