When a current-carrying wire has a current perpendicular to a magnetic field, it experiences a magnetic force proportional to its current, length, and the magnetic field. In other words, to quantify the force acting on the wire, the following formula can be used:
\[F = ILB\]
Where:
When we bend the wire into a loop and put it in a magnetic field, it can cause rotation, which is the same principle applied in electric motors (electromagnetic torque). We will discuss this more deeply in the following lessons.
A current-carrying wire of length m and current 6A is placed parallel to a magnetic field of strength 0.5T. Find the magnitude for the magnetic force applied on the wire.
Solution:
Since the wire is placed parallel to the magnetic field lines, there is no magnetic force applied to the wire. F = 0N
A 6m current-carrying wire is placed in a uniform magnetic field of strength 1T at an angle of 30 relative to the magnetic field lines and experiences a magnetic force with a magnitude of 24N. Find the magnitude of the current.
Solution:
\(F= IL \times B\)
\(I= \frac{F}{L \times B}\)
\(I= \frac{24}{6 \cdot 1 \cdot \sin(30)}\)
\(I=8 \: A\)
Written by Mubarak Aouda