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Physics · Ch 5 — Magnetism and Matter

Bar Magnet as an Equivalent Solenoid

5.2.2

Bar Magnet as an Equivalent Solenoid

The Core Idea: A Bar Magnet is Like a Solenoid

A current-carrying solenoid behaves like a magnetic dipole. The magnetic field lines of a bar magnet and a finite solenoid are strikingly similar, especially at large distances. This suggests that a bar magnet can be thought of as an equivalent solenoid — a collection of many tiny circulating currents (Ampere's hypothesis).

  • If you cut a bar magnet in half, you get two smaller magnets. Similarly, cutting a solenoid in half gives two smaller solenoids with weaker fields.
  • The field lines are continuous: they emerge from one face (the north pole) and enter the other face (the south pole).
  • A small compass needle shows the same deflection near a bar magnet and near a current-carrying solenoid, confirming the analogy.

Deriving the Axial Field of a Solenoid

To make the analogy quantitative, we calculate the magnetic field at a point PP on the axis of a finite solenoid (see Fig. 5.3(a) in the textbook). At large distances (r≫r \gg length of solenoid), the field simplifies to a form identical to that of a bar magnet.

The result is:

B=μ04π2mr3B = \frac{\mu_0}{4\pi} \frac{2m}{r^3}

Where:

  • BB = magnitude of the magnetic field at point PP on the axis, far from the solenoid.
  • μ0\mu_0 = permeability of free space (4π×10−7 T m/A4\pi \times 10^{-7} \, \text{T m/A}).
  • mm = magnetic moment of the solenoid (or the equivalent bar magnet).
  • rr = distance from the centre of the solenoid/magnet to point PP.

This is exactly the far axial field of a bar magnet (obtained experimentally). Therefore, a bar magnet and a solenoid produce the same magnetic field at large distances.

Key Result: Magnetic Moment Equivalence …

Figure 5.3Calculation of (a) The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet. (b) A magnetic needle in a uniform magnetic field B. The arrangement may be used to determine either B or the magnetic moment m of the needle.
Fig. 5.3 — Calculation of (a) The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet. (b) A magnetic needle in a uniform magnetic field B. The arrangement may be used to determine either B or the magnetic moment m of the needle.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Panel (a): Axial field of a finite solenoid

The left panel shows a finite solenoid — a cylindrical coil of wire — drawn in three dimensions. A dashed horizontal line runs through the centre of the solenoid, labelled O at the centre, and extends to a point P on the right. The solenoid has total length 2l2l, with half-length ll marked from O to the right end. Its radius is aa, shown as a vertical line from the axis to the top edge of the solenoid near a thin current-element ring.

A thin shaded ring of width dxdx is highlighted at a distance xx from O along the axis. This ring represents a single current loop of the solenoid. The distance from O to P is labelled rr. At P, a symbol ⊗\otimes indicates the magnetic field points into the page (the field direction along the axis of a solenoid).

The purpose of this panel is to set up the calculation of the axial magnetic field of a finite solenoid. By treating the solenoid as a stack of NN circular current loops, each of magnetic moment dm=(N dx/2l)⋅I⋅(πa2)d m = (N \, dx / 2l) \cdot I \cdot (\pi a^2), and integrating their contributions along the axis, one obtains the total field at P. At large distances (r≫l,ar \gg l, a), the result simplifies to the far-axial field of a bar magnet:

B=μ04π2mr3B = \frac{\mu_0}{4\pi} \frac{2m}{r^3}

where:

  • μ0\mu_0 is the permeability of free space,
  • mm is the magnetic moment of the solenoid (or bar magnet),
  • rr is the distance from the centre O to the point P.

This formula is equation (5.1) in the textbook. It shows that the solenoid behaves like a magnetic dipole at large distances, exactly as a bar magnet does.

Panel (b): Magnetic needle in a uniform field

The right panel shows a uniform magnetic field B\mathbf{B} represented by horizontal dashed lines with rightward arrowheads. A slim magnetic needle (a small bar magnet) is placed in this field, tilted at an angle θ\theta relative to the field direction. The north pole (N) is at the upper tip, the south pole (S) at the lower tip. A dashed reference line is drawn along the field direction, and the angle θ\theta is marked between the needle and this line.

This panel illustrates the torque experienced by a magnetic dipole (the needle) in a uniform external field. The torque tends to align the needle with the field. The magnitude of the torque is:

τ=mBsin⁡θ\tau = m B \sin\theta

where:

  • mm is the magnetic moment of the needle,
  • BB is the magnitude of the uniform field,
  • θ\theta is the angle between the needle’s axis and the field.

By measuring the deflection of the needle, one can determine either BB (if mm is known) or mm (if BB is known). This is the experimental arrangement described in the caption.

Physical idea …