A rectangular coil with 50 turns has side lengths of 0.12 meters and 0.1 meters. The current in the coil is three amps. Calculate the dipole moment of the coil when it is placed perpendicularly to a uniform magnetic field.
We can draw a sketch of this coil with its dimensions drawn in. And we know that if we were looking at it side on, we would see that it has 50 turns to this rectangular frame. The current running through this coil is three amps. And we’re told that a uniform magnetic field is oriented perpendicular to this coil. In this case, we’ve drawn it going into the page.
If we look up the mathematical relationship for the magnitude of a dipole moment of a current-carrying wire, we see that it’s equal to the current running through the coil times its cross-sectional area multiplied by the number of turns in the coil all times the sin of the angle 𝜃, where 𝜃 is the angle between the plane of the coil and the applied magnetic field.
In our case, since the plane of our coil is perpendicular to the applied field, we know that 𝜃 is 90 degrees and that the sin of 90 degrees is one. Our dipole moment then equals 𝐼 times 𝐴 times 𝑁, or three amperes times 0.1 meters times 0.12 meters, the area of our coil, multiplied by 50. Multiplying these values together, we find a result of 1.8 amps meters squared. That’s the dipole moment of this coil when it’s placed in a perpendicular magnetic field.