Current-carrying Solenoid: An Electromagnetic Equivalent of a Bar-magnet

A solenoid is a long cylindrical coil made by winding many turns of insulated wire closely together.

When electric current flows through the coil, every circular turn produces a magnetic field. The fields due to all turns combine to produce a strong magnetic field along the axis of the solenoid.

Think of it as many tiny current loops working together like one large magnet.

Suspend a current-carrying solenoid freely using a long thread. It turns and finally comes to rest approximately along the north–south direction.

This happens because the solenoid behaves like a magnetic dipole:

• The end that points geographically north is called its north pole.
• The opposite end is called its south pole.

A freely suspended bar magnet behaves in the same way. Therefore, a current-carrying solenoid has magnetic behaviour similar to a bar magnet.

Bring the end of a second current-carrying solenoid close to one end of a freely suspended solenoid.

Thus, current-carrying solenoids obey the same pole rule as bar magnets: Like poles repel; unlike poles attract.

The magnetic field pattern of a current-carrying solenoid is very similar to that of a bar magnet.

• Outside the solenoid, field lines emerge from the north pole and enter the south pole.
• Inside the solenoid, field lines travel from south to north.
• Hence, magnetic field lines always form continuous closed loops.
• For a long solenoid, the field lines inside are nearly parallel and equally spaced. This shows that the magnetic field inside is nearly uniform.

For a solenoid:

• N = total number of turns
• I = current through the coil
• A = cross-sectional area of each turn

The magnetic dipole moment is: m = NIA

For a solenoid of length 2l, having n turns per unit length, N = n(2l)

If its radius is a, then A = πa²

Therefore, m = (n 2l) I (πa²)

SI unit of magnetic dipole moment: A m²

Consider a solenoid of:

• radius a,
• length 2l,
• turns per unit length n,
• current I.

Let P be a point on its axis at distance r from its centre.

Take a thin part of the solenoid of thickness dx, located at distance x from the centre. It contains: ndx turns.

The magnetic field at P due to this thin element is:

• dB = \[\frac{\mu_0}{4\pi}\frac{2(ndx)Ia^2}{[a^2+(r-x)^2]^{3/2}}\]

Adding the contributions of all elements from x = −l to x = +l,

• B = \[\frac{\mu_0}{4\pi}2nIa^2\int_{-l}^{+l}\frac{dx}{[a^2+(r-x)^2]^{3/2}}\]

For a point far away from the solenoid, r ≫ a and r ≫ 2l

• so that [a² + (r − x)²]^3/2 ≈ r³

Therefore,

• B ≈ \[\frac{\mu_0}{4\pi}\frac{2nIa^2}{r^3}\int_{-l}^{+l}dx\]

Since

• \[\int_{-l}^{+l}dx=2l\]

we obtain

• B = \[\frac{\mu_0}{4\pi}\frac{2(n2l)I(\pi a^2)}{r^3}\]

But m = (n 2l) I (πa²)

Hence, 

• B = \[\frac{\mu_0}{4\pi}\frac{2m}{r^3}\]

This is exactly the expression for the magnetic field on the axial line of a bar magnet at a large distance.

• A current-carrying solenoid behaves like a magnetic dipole.
• Like poles repel and unlike poles attract.
• Anticlockwise current at an observed end means that end is north.
• Clockwise current at an observed end means that end is south.
• Magnetic dipole moment of a solenoid is: m = NIA
• Far axial field of a solenoid is: B = \[\frac {μ_0}{4π}\frac {2m}{r^3}\]
• This is the same far-axial field as that of a bar magnet.