Magnetic Force on a Current-Carrying Wire Calculator
Find the force on a straight current-carrying wire in a magnetic field, per metre of wire, with the right hand rule direction.
What it does
Put a current-carrying wire in a magnetic field and it moves. That is the whole of the electric motor, the coil in a loudspeaker, the needle of a meter and the relay in a starter, and the force behind all of them comes from one short formula.
The OpenStax College Physics text derives it by starting with a single charge. The force on a charge moving through a field is q times v times B times the sine of the angle, and the wire is made of a great many charges all drifting the same way, so the forces add. Collecting the terms gives the current flowing through the wire, and the result is F equals I l B sin of theta.
The same source is careful about the direction. The force is given by the right hand rule with the thumb along the current, and dividing the force by the length gives the force per unit length, which is the form that appears most often in practice. The LibreTexts version of the same derivation points out that the force is the net force on the charge carriers themselves, and that the conductor feels it because the charges cannot escape.
This page runs that calculation. Give it the current, the length of wire in the field, the field strength and the angle between the current and the field, and it returns the force, the force per unit length, the force the wire would feel at right angles and the component of the field that acts across it.
The limits are worth stating. The formula is for a straight wire in a uniform field, so a coil has to be treated as a series of straight pieces, and a field that varies along the wire needs the pieces treated separately. It also assumes the wire is light enough that the force does not move it, which in a motor is exactly what does not happen.
How to use
- Choose metric or imperial for the length, the field and the answer.
- Enter the current running in the wire.
- Enter the length of wire that sits in the field. Only that part of the wire feels the force.
- Enter the magnetic field strength. Use 1.5 T for the field near a decent permanent magnet and 0.0001 T for the Earth's field.
- Enter the angle between the current and the field, with 90 degrees for a wire square to the field.
- Read the force and the force per unit length, and use the right hand rule direction to work out which way the wire moves.
Formula
- F
- Force on the wire, in newtons.
- I
- Current flowing through the wire, in amperes.
- l
- Length of wire that sits in the field, in metres.
- B
- Magnetic field strength, in tesla.
- theta
- Angle between the direction of the current and the direction of the field, in degrees.
The sine of the angle is what makes the force vanish for a wire parallel to the field, since there is then no component of the field across the wire. The field component that actually acts is B times the sine of theta, and the force per unit length is the number that sizes a motor.
Worked example
Example: 20 A through 50 mm of wire in a field of 1.5 T
Defaults: metric units, a current of 20 A, a field of 1.5 T, a length of 0.05 m and an angle of 90 degrees.
- The length in metres is 0.05 and the field is 1.5 T.
- The sine of 90 degrees is 1, so the angle costs nothing here.
- The force is 20 x 0.05 x 1.5 x 1 = 1.5000 N.
- The force per unit length is 1.5000 over 0.05 = 30.0000 N per m, which is also I times B times the sine of theta.
- The component of the field across the wire is 1.5 times the sine of 90 degrees, which is 1.500000 T.
- At right angles the force is the same, because the wire is already square to the field, so the maximum is 1.5000 N.
- The tool shows 1.5000 N, 30.0000 N per m, 1.5000 N, 1.500000 T and a force pointing to the left of the current by the right hand rule.
Reference table
| Angle | Field across the wire | Force per metre | Force on 50 mm |
|---|---|---|---|
| 0 degrees | 0.000000 T | 0.0000 N per m | 0.0000 N |
| 30 degrees | 0.750000 T | 15.0000 N per m | 0.7500 N |
| 60 degrees | 1.299038 T | 25.9808 N per m | 1.2990 N |
| 90 degrees | 1.500000 T | 30.0000 N per m | 1.5000 N |
Limits
- The formula is for a straight wire in a uniform field, so a curved wire has to be split into short straight pieces and the results added.
- It assumes the wire does not move far enough to change the field or the geometry, which in a motor is exactly what happens.
- It says nothing about the force on the individual charges inside the wire, only on the wire as a whole.
- There is no allowance for a field that changes along the length of the wire, or for a wire that is not straight.
Common mistakes
- Using the angle between the current and the field when you mean the angle between the wire and the field, which is what the formula actually needs.
- Forgetting the sine, so a wire lying along the field is given a full force when it should have none.
- Mixing up tesla and gauss, which differ by a factor of 10000.
- Using the whole length of the wire when only the part inside the field feels the force.
- Forgetting the right hand rule and reporting a magnitude with no idea which way the wire moves.
Try it yourself
Quick checks. Answers sit tucked under each question.
What happens if the angle drops to 30 degrees?
The force per metre falls to 15.0000 N per m, exactly half the value at right angles, and the force on the 50 mm piece becomes 0.7500 N. The field component that acts across the wire is 0.750000 T instead of 1.500000 T.
How much force does a 2 m length of the same wire feel?
30.0000 N, which is forty times the 1.5000 N of the 50 mm piece, because the force is proportional to the length of wire in the field. The force per metre stays at 30.0000 N per m however long the wire is.
What if the current is doubled?
The force doubles, because the current is in the formula directly. At 40 A the 50 mm of wire would feel 3.0000 N, and the force per metre would be 60.0000 N per m. The same holds for the field, and neither of them is squared anywhere.
Questions
What is the force on a current-carrying wire?
It is the current times the length of wire in the field times the field strength times the sine of the angle between the current and the field. The OpenStax text calls it the equation for the magnetic force on a length of wire carrying a current in a uniform magnetic field.
Why does the force per unit length matter more?
Because a coil of many turns is a set of short straight pieces, and the force on each piece is proportional to its length. The force per unit length is the number that decides how much torque a coil produces.
How do I find the direction?
The right hand rule. Point the thumb of your right hand along the current and curl your fingers towards the field, and your palm pushes out in the direction the force goes. The OpenStax text says the same thing in words.
What if the wire is parallel to the field?
The force is zero, because the sine of zero degrees is zero. There is no component of the field across the wire, so the charges feel nothing sideways.
Does the force depend on the material of the wire?
Not in this formula. The material shows up in a different place, through the current the wire carries for a given voltage and through the strength of its own magnetic field.
What is the field component across the wire?
The field multiplied by the sine of the angle, which is the part of the field that pushes sideways on the charges. It is the only part of the field that does any work here.
Sources
- OpenStax College Physics 2e, 22.7: magnetic force on a current-carrying conductor
- Physics LibreTexts, 11.5: magnetic force on a current-carrying conductor, with the worked example