# 25.2.1 Force on a Current-Carrying Conductor

### Force on a Current-Carrying Conductor

• A current-carrying conductor produces its own magnetic field
• When interacting with an external magnetic field, it will experience a force
• A current-carrying conductor will only experience a force if the current through it is perpendicular to the direction of the magnetic field lines
• A simple situation would be a copper rod placed within a uniform magnetic field
• When current is passed through the copper rod, it experiences a force which makes it move

### Calculating Magnetic Force on a Current-Carrying Conductor

• The strength of a magnetic field is known as the magnetic flux density, B
• This is also known as the magnetic field strength
• It is measured in units of Tesla (T)
• The force F on a conductor carrying current I at right angles to a magnetic field with flux density B is defined by the equation

F = BIL sinθ

• Where:
• F = force on a current carrying conductor in a B field (N)
• B = magnetic flux density of external B field (T)
• I = current in the conductor (A)
• L = length of the conductor (m)
• θ = angle between the conductor and external B field (degrees)
• This equation shows that the greater the current or the magnetic field strength, the greater the force on the conductor
• The maximum force occurs when sin θ = 1
• This means θ = 90o and the conductor is perpendicular to the B field
• This equation for the magnetic force now becomes:

F = BIL

• The minimum force (0) is when sin θ = 0
• This means θ = 0o and the conductor is parallel to the B field
• It is important to note that a current-carrying conductor will experience no force in a magnetic field if is parallel to the field

#### Worked example

Step 1:            Write down the known quantities

Magnetic flux density, B = 80 mT = 80 × 10-3 T

Current, I = 0.87 A

Length of wire, L = 1.4 m

Angle between the wire and the magnetic field, θ = 30o

Step 2:            Write down the equation for force on a current-carrying conductor

F = BIL sinθ

Step 3:            Substitute in values and calculate

F =  (80 × 10-3) × (0.87) × (1.4) × sin(30) = 0.04872 = 0.049 N (2 s.f)

#### Exam Tip

Remember that the direction of current flow is the flow of positive charge (positive to negative), and this is in the opposite direction to the flow of electrons

### Author: Katie

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.
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