Current Loop as a Magnetic Dipole (original) (raw)

Last Updated : 1 Jun, 2026

When an electric charge moves, it produces an electric current and creates a magnetic field around it. This magnetic field can exert forces on nearby moving charges and magnets. Stationary charges do not produce magnetism, while moving charges can cause attraction or repulsion.

Magnets and Magnetic fields

A moving charge creates a magnetic field, and the force experienced in a magnetic field is called the magnetic force. Charge is a fundamental property of matter that allows it to produce and experience electrical and magnetic effects. The region in space around a magnet where its magnetic influence is felt is called the magnetic field of the magnet.

Magnetic fields of a magnet

If a point charge q moves with velocity v at position r in the presence of both an electric field Er and a magnetic field Br​, the total force on the charge is given by:

F = q [Er + v × Br] = EElectric + Fmagnetic

This formula, known as the Lorentz force, was stated by H. A. Lorentz based on the experiments of Ampere and others.

**Characteristics of Magnets

Magnetic Dipole and Magnetic Dipole Moment

A magnetic dipole consists of two equal and opposite magnetic poles separated by a small distance. A current-carrying loop behaves like a magnetic dipole and produces a magnetic field similar to a bar magnet.

The magnetic dipole moment is a vector quantity that measures the strength and direction of a magnetic dipole. Its direction is from south to north, and its SI unit is A·m².

The torque on a magnetic dipole is:

\tau = mB \sin\theta

Where,

A magnetic dipole consists of two equal and opposite magnetic poles (north and south), which together produce a magnetic field. **Example: when a bar magnet is broken into smaller pieces, each piece behaves as an independent magnetic dipole, with its own north and south poles.

Magnetic Dipole Moment of a Circular Loop

When an electric current flows through a circular loop, it behaves like a magnetic dipole. This happens because the moving charges in the loop generate a magnetic field, similar to that produced by a bar magnet. The loop tends to align itself in an external magnetic field, just like a magnetic dipole.

The magnetic dipole moment of a current-carrying loop is defined as the product of the current flowing through the loop and the area enclosed by it. For a coil having multiple turns, the dipole moment increases proportionally with the number of turns.

μ = nIA

Where,

Current loop Magnetic Field

The SI unit of magnetic dipole moment is ampere–meter² (A·m²), while in the CGS system, it is expressed in erg per gauss. Here, _erg represents the unit of energy and _gauss represents the unit of magnetic flux density.

**Characteristics

Current Loop as a Magnetic Dipole

A circular current-carrying loop behaves like a magnetic dipole because the flow of current in a closed loop produces a magnetic field similar to that of a bar magnet. The direction of this magnetic field is determined by the right-hand thumb rule. Due to this, the loop exhibits north and south pole behavior, making it equivalent to a magnetic dipole.

The magnetic dipole moment of a circular current loop is given by:

\mu = n i \pi r^2

Where,

Current Loop as a Magnetic Dipole Derivation

Current Loop as a Magnetic Dipole

Consider a circular loop of radius R placed on a table, carrying a current i in the anti-clockwise direction. Let a point P be located on the axis of the loop at a distance l from its centre. The magnetic field at point P is given by:

B = \frac{\mu_0 i R^2}{2(R^2 + l^2)^{3/2}}

For simplification, assume that the point P is very far from the loop (l >> R). Then, the magnetic field can be approximated as:

B \approx \frac{\mu_0 i R^2}{2 l^3} = \frac{\mu_0}{4 \pi} \frac{2 i (\pi R^2)}{l^3}

The area of the loop is:

A = \pi R^2

Thus, the magnetic field can be expressed in terms of the magnetic dipole moment μ as:

B = \frac{\mu_0}{4 \pi} \frac{2 i A}{l^3} = \frac{\mu_0}{4 \pi} \frac{2 \mu}{l^3}

where μ=iA is the magnetic dipole moment of the current loop.

This formula for the magnetic field of a current loop is analogous to the electric field of a dipole:

\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{2 \vec{p}}{r^3}

Here, \vec{p} is the electric dipole moment and r is the distance from the dipole.

Solved Problems

**Question 1: Two wires of the same length are shaped into a square and a circle. If they carry the same current, then the ratio of their magnetic moments is:

**Solution: The magnetic moment of a loop is given by

m = I \cdot A

Let the side of the square be a and the radius of the circle be r. Since the wires have the same length

\text{Perimeter of square} = \text{Circumference of circle} \implies 4a = 2 \pi r \implies a = \frac{\pi r}{2} \quad ...(1)

The area of the square loop:

A_1 = a^2 = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4}

Thus, the magnetic moment of the square loop:

m_1 = I \cdot A_1 = I \frac{\pi^2 r^2}{4} \quad ...(2)

The area of the circular loop:

A_2 = \pi r^2

Thus, the magnetic moment of the circular loop:

m_2 = I \cdot A_2 = I \pi r^2 \quad ...(3)

The ratio of magnetic moments:

\frac{m_1}{m_2} = \frac{I \frac{\pi^2 r^2}{4}}{I \pi r^2} = \frac{\pi}{4}

\frac{m_1}{m_2} = \frac{\pi}{4}

**Question 2: A circular coil of 300 turns and a diameter of 14 cm carries a current of 15 A. The magnitude of the magnetic moment associated with the loop is going to be

**Solution: Given:

Number of turns (N) = 300

Radius of coil (r) = 14/2 = 7cm = 7 × 10-2 m

Current in coil (I) = 15A

Magnetic moment of a circular coil:

M = N I A = N I (\pi r^2)

Substitute the values:

M = 300 \times 15 \times \pi \times (0.07)^2

M \approx 69.2 \text{ Am}^2

**Question 3: A circular coil of 100 turns, each turn of radius 8 cm carries a current of 0.4 A. What is the magnitude of the magnetic field B at the center of the coil?

**Solution: Given:

Number of turns (N) = 100

radius of each turn (r) = 8cm = 8 × 10-2m

Current flowing in the coil (I) = 0.4 A

permeability of free space (μ0) = 4π × 10-7 TmA-1

The magnetic field at the center of a circular coil is given by:

B = \frac{\mu_0 N I}{2 r}

Substitute the values:

B = \frac{4 \pi \times 10^{-7} \times 100 \times 0.4}{2 \times 0.08}

B \approx 3.14 \times 10^{-4} \, \text{T}

**Question 4: A long straight wire in the horizontal plane carries 50 A in the north-to-south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

**Solution: Given:

Current in the wire (I) = 50 A

Distance of a point from the wire (r) = 2.5 m

permeability of free space (μ0) = 4π × 10-7 TmA-1

The magnetic field due to a long straight wire is:

B = \frac{\mu_0 I}{2 \pi r}

Substitute the values:

B = \frac{4 \pi \times 10^{-7} \times 50}{2 \pi \times 2.5} = 8 \times 10^{-6} \, \text{T}

Using the right-hand thumb rule, the magnetic field at a point east of the wire (current north-to-south) points into the page.

B = 8 \times 10^{-6} T, into the page

Unsolved Problems

**Question 1: A circular coil of 50 turns has a radius of 10 cm. It carries a current of 2 A. Find its magnetic moment.

Question 2: A straight wire 1 m long carries a current of 5 A. It is placed in a uniform magnetic field of 0.1 T perpendicular to the wire. Find the force on the wire.

**Question 3: Two long parallel wires, 1 m apart, carry currents of 6 A and 9 A in the same direction. Find the force per meter between them.

**Question 4: Two wires of equal length are shaped into a square and an equilateral triangle. Both carry the same current. Find the ratio of their magnetic dipole moments.

**Question 5: A circular coil of 300 turns and radius 8 cm carries a current of 0.5 A. Find the magnetic field at a point on its axis, 20 cm from its centre.