Question

Derive an expression for the axial magnetic field of a finite solenoid of length 2l and radius r carrying current I. Under what condition does the field become equivalent to that produced by a bar magnet?

Answer 1

Consider a solenoid of length 2l, radius r and carrying a current I and having n turns per unit length.

Consider a point P at a distance a from the centre O of solenoid. Consider an element of solenoid of length dx at a distance x from its centre. This element is a circular current loop having (ndx) turns. The magnetic field at axial point P due to this current loop is

The total magnetic field due to entire solenoid is

For a>>I and a>>r, we have {r²+(a - x)²}^3/2 = a³

The magnetic moment of solenoid m = (= NIA) = (n2I)I. πr²

This is also the far axial magnetic field of a bar magnet. Hence, the magnetic field, due to current carrying solenoid along its axial line is similar to that of a bar magnet for far off axial points.