The Bar Magnet as an Equivalent Solenoid

Ampere's Hypothesis: All Magnetism Is Circulating Currents

In Chapter 4 (Section 4.9) we saw that a current loop acts as a magnetic dipole, with magnetic moment

$$m = NIA$$

where $N$ is the number of turns, $I$ the current and $A$ the loop area. Ampere's hypothesis takes this idea all the way: all magnetic phenomena can be explained in terms of circulating currents. A bar magnet, on this view, is nothing mysterious — it is a large number of tiny circulating currents at the atomic level, stacked together just like the turns of a solenoid.

The evidence for the analogy:

• The field-line patterns of a bar magnet and a finite solenoid are strikingly similar (Section 1).
• Cutting a bar magnet in half is like cutting a solenoid — you get two smaller solenoids with weaker magnetic properties. The field lines remain continuous, emerging from one face and entering the other. This neatly explains why you can never isolate a pole!
• Move a small compass needle around a bar magnet and around a current-carrying finite solenoid: the deflections are similar in both cases.

Key Point: Poles are a convenient model; the deeper source of magnetism is current. The 'north face' of a solenoid is simply the face where the current appears anticlockwise (recall the clock rule from Chapter 4).

The magnetic moment vector $\vec{m}$ of a bar magnet points from its south pole to its north pole inside the magnet — exactly like the area vector of the equivalent current loops.

The Axial Field of a Finite Solenoid

To make the analogy firm, let's compute the field of a finite solenoid at a faraway point P on its axis and show that it matches a bar magnet's far axial field.

Setup: a solenoid of radius $a$, length $2l$, with $n$ turns per unit length carrying current $I$. P lies on the axis at distance $r$ from the centre, with $r \gg a$ and $r \gg l$.

The result:

$$B = \frac{\mu_0}{4\pi}\frac{2m}{r^3}$$

where the total magnetic moment is $m = n(2l) \cdot I \cdot \pi a^2$ — i.e. (total turns) $\times$ (current) $\times$ (area), which is just $m = NIA$ again.

[JEE Tip] The derivation (good for building skill, though exams need only the final result): treat a slice of thickness $dx$ at distance $x$ from the centre as a circular loop of $n\,dx$ turns. Its axial field at P is $dB = \frac{\mu_0 (n\,dx) I a^2}{2[(r-x)^2 + a^2]^{3/2}}$. For $r \gg a, l$ the denominator is approximately $r^3$, so integrating $x$ from $-l$ to $+l$ gives $B = \frac{\mu_0 n I (2l) a^2}{2 r^3} = \frac{\mu_0}{4\pi}\frac{2m}{r^3}$.

This is also the far axial field of a bar magnet measured experimentally. Same far-field formula, same $1/r^3$ fall-off, and in the far-field dipole approximation, the same dipole field pattern: the bar magnet and the solenoid are equivalent magnetic dipoles.

What 'Equivalent' Really Means

The equivalence is quantitative, not just visual:

The magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same far-field magnetic dipole behaviour.

Three practical consequences:

1. You can measure $m$ of a short magnet from its far axial field. Measure $B$ on the axis at a large distance $r$; then $m = \frac{4\pi}{\mu_0}\frac{B r^3}{2}$.
2. The $1/r^3$ signature. A dipole field falls off as $1/r^3$ — much faster than the $1/r^2$ of a point charge's electric field. Double the distance and the axial field drops by a factor of 8.
3. Cutting rules follow automatically. Cut a bar magnet transverse to its length: each half is a solenoid with half the turns, so $m$ halves. Cut it along its length: each half has half the loop area, so $m$ halves again. Either way you always get complete, weaker magnets.

[NEET Important] The SI unit of magnetic moment is A m$^2$ (ampere metre squared), equivalently J/T (joule per tesla) — both appear in NEET options. Verify: $U = -mB\cos\theta$ has joules when $m$ is in J/T and $B$ in tesla, and $m = NIA$ has A m$^2$. They are the same unit.

Useful constant for numericals: $\frac{\mu_0}{4\pi} = 10^{-7}$ T m A$^{-1}$.

Solved Examples

Example 1: Magnetic moment of a solenoid

A closely wound solenoid of 2000 turns and cross-sectional area $1.6 \times 10^{-4}$ m$^2$ carries a current of 4.0 A. What is its magnetic moment?

Solution:

1. Formula: $m = NIA$.
2. Substitute: $m = 2000 \times 4.0 \times 1.6 \times 10^{-4}$.
3. Calculate: $m = 2000 \times 4.0 = 8000$; $8000 \times 1.6 \times 10^{-4} = 1.28$.
4. Answer: $m = 1.28$ A m$^2$, directed along the solenoid axis (S face to N face).

Takeaway: This solenoid behaves like a bar magnet of moment 1.28 J/T in the dipole approximation.

Example 2: Far axial field of the same solenoid

Find the magnetic field on the axis of the above solenoid ($m = 1.28$ A m$^2$) at a distance of 20 cm from its centre (far compared to its size).

Solution:

1. Formula: $B = \frac{\mu_0}{4\pi}\frac{2m}{r^3}$.
2. Substitute: $r = 0.2$ m, so $r^3 = 8 \times 10^{-3}$ m$^3$. $B = 10^{-7} \times \frac{2 \times 1.28}{8 \times 10^{-3}}$.
3. Calculate: $\frac{2.56}{8 \times 10^{-3}} = 320$; $B = 10^{-7} \times 320 = 3.2 \times 10^{-5}$ T.
4. Answer: $B = 3.2 \times 10^{-5}$ T along the axis.

Example 3: Cutting a solenoid (and a magnet)

A bar magnet of moment $m$ is cut into two equal halves (i) transverse to its length, (ii) along its length. What is the moment of each piece?

Solution:

1. Model the magnet as a solenoid of N turns, current I, area A, with $m = NIA$.
2. (i) Transverse cut: each half is a solenoid of $N/2$ turns, same $I$, same $A$: $m' = (N/2)IA = m/2$.
3. (ii) Lengthwise cut: each half keeps all N 'turns' but with half the cross-sectional area: $m' = NI(A/2) = m/2$.
4. Answer: $m/2$ in both cases — and each piece is still a complete magnet.

Example 4: The $1/r^3$ fall-off

The axial field of a short bar magnet at 10 cm from its centre is $B_0$. What is it at 20 cm?

Solution:

1. For a dipole, $B \propto 1/r^3$ on the axis (far field).
2. Distance doubles: $r \to 2r$, so $B \to B_0/2^3 = B_0/8$.
3. Answer: $B_0/8$. Doubling distance costs a factor of 8 — the dipole signature.

Example 5: Measuring a magnet's moment from its field

The magnetic field on the axis of a short bar magnet, 0.5 m from its centre, is measured to be $1.6 \times 10^{-7}$ T. Find the magnetic moment of the magnet.

Solution:

1. Formula (rearranged): from $B = \frac{\mu_0}{4\pi}\frac{2m}{r^3}$, we get $m = \frac{B r^3}{2 \times 10^{-7}}$.
2. Substitute: $r^3 = 0.125$ m$^3$, so $m = \frac{1.6 \times 10^{-7} \times 0.125}{2 \times 10^{-7}}$.
3. Calculate: numerator $= 2 \times 10^{-8}$; dividing by $2 \times 10^{-7}$ gives $m = 0.1$.
4. Answer: $m = 0.1$ A m$^2$. This is how a magnet's moment can be determined experimentally from its far axial field.

Example 6: Doubling turns and current

A solenoid's number of turns is doubled and its current halved, keeping geometry fixed. What happens to (a) its magnetic moment, (b) its far axial field at a fixed point?

Solution:

1. $m = NIA$: with $N \to 2N$ and $I \to I/2$, $m \to (2N)(I/2)A = NIA = m$ — unchanged.
2. $B = \frac{\mu_0}{4\pi}\frac{2m}{r^3}$ depends only on $m$ and $r$, so $B$ is unchanged too.
3. Takeaway: the far field cares only about the product $NIA$, not the individual factors.

Example 7: Moment from solenoid geometry

A solenoid has 500 turns per metre, length 10 cm, radius 1.0 cm, and carries 2.0 A. Treating it as a bar magnet, find its magnetic moment.

Solution:

1. Total turns: $N = n \times \text{length} = 500 \times 0.10 = 50$ turns.
2. Area: $A = \pi a^2 = 3.14 \times (0.01)^2 = 3.14 \times 10^{-4}$ m$^2$.
3. Moment: $m = NIA = 50 \times 2.0 \times 3.14 \times 10^{-4} = 3.14 \times 10^{-2}$ A m$^2$.
4. Answer: $m \approx 0.031$ A m$^2$ along the axis. This is the $m = n(2l)I\pi a^2$ of the equivalent-solenoid derivation.

Example 8: Same far-field deflection test

A compass needle is moved in the far-field region around box X and box Y. X contains a short bar magnet; Y contains an equivalent current-carrying solenoid with the same magnetic moment. Can the compass tell which is which from the far-field pattern alone?

Solution:

1. The far field of a bar magnet and that of an equivalent solenoid are the same dipole field.
2. The compass responds only to the local magnetic field, so in the far-field dipole approximation the deflection patterns are the same.
3. Answer: No — from far-field compass measurements alone, X and Y are indistinguishable. Very close to the objects, detailed fields may depend on geometry, but the equivalence meant here is the dipole/far-field one.

Example 9: Ratio of axial fields of two magnets

Two short bar magnets have moments $m$ and $2m$. At what distance ratio do they produce equal axial fields?

Solution:

1. Equal fields: $\frac{\mu_0}{4\pi}\frac{2m}{r_1^3} = \frac{\mu_0}{4\pi}\frac{2(2m)}{r_2^3}$.
2. So $\frac{r_2^3}{r_1^3} = 2$, i.e. $\frac{r_2}{r_1} = 2^{1/3} \approx 1.26$.
3. Answer: the stronger magnet produces the same field about 26% farther away.

Example 10: Why the lines stay continuous when you cut

When a solenoid is cut into two, why do the field lines remain continuous closed loops through each piece?

Solution:

1. Each piece is itself a stack of complete current loops — a smaller solenoid.
2. Field lines of any current distribution form closed loops; there are no magnetic monopoles to start or end them.
3. So each half develops its own N and S faces, with lines emerging from one face, looping around, entering the other face and continuing inside. The same logic explains why cutting a bar magnet never isolates a pole.