A 20-centimeter section of wire carrying a current of 12 amperes is positioned at 90 degrees to a 0.1-tesla magnetic field. What is the size of the force acting on the wire?
Let’s say that this is our section of wire. It has a length 𝐿 of 20 centimeters. It carries a current 𝐼 of 12 amperes. And it exists in a magnetic field 𝐵 of strength 0.1 teslas that is perpendicular to the axis of the wire. Knowing all this, we want to solve for the force, we’ll call it 𝐹, that acts on the wire. We can recall that the force on a current-carrying wire of length 𝐿 carrying a current 𝐼 in a magnetic field of strength 𝐵 when the current 𝐼 in the magnetic field 𝐵 are perpendicular to one another is equal to 𝐵 times 𝐼 times 𝐿. This equation applies to our scenario where indeed the wire and the magnetic field are perpendicular to one another. In our case, the magnetic field 𝐵 is 0.1 teslas, the current 𝐼 is 12 amperes, and the length 𝐿 is 20 centimeters.
Before we calculate this force, we’ll want to convert the units of centimeters into units of meters. This way, all of our units will be written according to the same SI standard. One centimeter, we recall, is equal to 10 to the negative two or one one hundredth of a meter. This means that 20 centimeters equals 20 times 10 to the negative two meters or, written another way, 0.20 meters. When we multiply these three quantities together, we get a result of 0.24 with units of newtons. This is the force that will be experienced by this 20-centimeter section of wire.