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Magnetic Force on Current

F = IL cross B, force between parallel conductors.

Darshan N
Updated: 19 March 2026
10 min read

When a current-carrying conductor is placed in an external magnetic field, it experiences a mechanical force. This principle is the operating basis of electric motors, galvanometers, relays, and loudspeakers. The ability to convert electrical energy into mechanical motion through this force is one of the most practically significant results in electromagnetic theory.

Force on Current-Carrying Conductor in Magnetic FieldConductor (I)I (up)B (into page shownsymbolically)F = IL x B(force directionby right-hand rule)Key FormulasElement force:dF = I dL x BStraight conductor:F = IL x BMagnitude:|F| = BIL sin(theta)Parallel wires:F/L = mu_0 I1 I2 / 2pi d
Figure 1: A current I in conductor of length L placed in field B experiences force F = IL x B perpendicular to both.

Force on a Current Element in a Magnetic Field

The magnetic force on a current element is expressed as dF = I dL x B, where dL is a differential length vector in the direction of current flow and B is the external magnetic flux density. For a straight conductor of length L carrying current I in a uniform field B, the total force is F = IL x B. The magnitude is |F| = BIL sin(theta), where theta is the angle between the direction of current and B.

The direction of the force is always perpendicular to both the current direction and B, determined by the right-hand rule or the cross-product rule. When the current is parallel to B (theta = 0), the force is zero. When the current is perpendicular to B (theta = 90 degrees), the force is maximum and equals BIL. This geometry is used in the design of DC motors where armature conductors are arranged to remain perpendicular to the field for maximum torque.

Force Between Two Parallel Current-Carrying Conductors

Two parallel conductors carrying currents I1 and I2 separated by distance d exert forces on each other via their respective magnetic fields. The force per unit length between them is given by F/L = mu_0 * I1 * I2 / (2*pi*d). If the currents flow in the same direction, the conductors attract each other. If they flow in opposite directions, they repel. This result is historically important: it forms the basis of the SI definition of the Ampere.

In practical wiring and busbar design, large parallel conductors carrying high currents in the same direction will experience attractive forces. Engineers must mechanically support busbars to withstand these forces, especially during fault conditions when currents can be very large. This is a direct engineering consequence of F = IL x B.

Practical Understanding

In a galvanometer, a rectangular current-carrying coil is placed in a radial magnetic field so that the coil sides always remain perpendicular to B regardless of rotation angle. This ensures the torque on the coil is proportional only to current and not to angle, giving a linear scale. The torque on one side of the coil of width w carrying current I in field B is tau = BIlw = BINA for N turns and area A, where this torque is balanced by a spring to give deflection.

Example
Given:
Two long parallel conductors separated by d = 5 cm = 0.05 m.
Current I1 = 20 A, I2 = 30 A, both in same direction.
Length of each conductor L = 2 m.

Why this formula applies:
Each conductor lies in the magnetic field produced by the other.
Field from conductor 1 at conductor 2: B1 = mu_0*I1 / (2*pi*d)
Force on conductor 2: F = I2 * L * B1

Formula:
F = mu_0 * I1 * I2 * L / (2*pi*d)

Substitution:
F = (4*pi*1e-7 * 20 * 30 * 2) / (2*pi*0.05)

Calculation:
Numerator: 4*pi*1e-7 * 1200 = 1.5079e-3
Denominator: 2*pi*0.05 = 0.3142
F = 1.5079e-3 / 0.3142 = 4.8e-3 N

Final Answer:
F = 4.8 mN  (attractive, since currents are in same direction)
Exam Tip: For parallel conductors, F/L = mu_0*I1*I2 / (2*pi*d). Same-direction currents attract; opposite-direction currents repel. Maximum force on a single conductor in field B is when current is perpendicular to B giving |F| = BIL.
Force Between Two Parallel ConductorsWire 1I1 upWire 2I2 upF1F2d(separation)Same direction I:Conductors ATTRACTOpposite direction I:Conductors REPELF/L = mu_0 I1 I2----------------- 2 pi d
Figure 2: Parallel conductors with same-direction currents attract each other. Force per unit length F/L = mu_0*I1*I2 / (2*pi*d).
  • The force on a current element dF = I dL x B; for uniform field and straight conductor F = IL x B.
  • Magnitude |F| = BIL sin(theta); maximum when theta = 90 degrees, zero when theta = 0 degrees.
  • Parallel conductors with same-direction currents attract; opposite-direction currents repel.
  • Force per unit length between parallel conductors: F/L = mu_0 * I1 * I2 / (2*pi*d).
  • This principle drives motors (torque via F = IL x B) and galvanometers (proportional deflection in radial field).

Quick Revision

  • dF = I dL x B; for straight wire in uniform field F = IL x B.
  • |F| = BIL sin(theta); maximum force at theta = 90 degrees.
  • F/L = mu_0*I1*I2 / (2*pi*d) for two parallel conductors.
  • Same direction: attraction. Opposite direction: repulsion.
  • Trap: F = BIL applies only when B is perpendicular to I; otherwise multiply by sin(theta).
  • SI definition of Ampere is based on force between parallel conductors.
  • Direction of force: always perpendicular to both I and B, use right-hand cross-product rule.

Force on Current

Test your ability to compute magnetic forces on current-carrying conductors and between parallel wires.

Question 1 of 3

Q1.A straight conductor of length L carrying current I is placed in a uniform magnetic flux density B. The force on the conductor is: