# Question Video: Force on an Electrical Conductor in a Magnetic Field

A 50.0 cm length of wire is in a 1.20-T magnetic field. The current in the wire is 8.00 A and a force of 2.40 N acts on the wire. What is the angle between the wire and the magnetic field? If the wire is rotated to be at an angle of 90.0° to the magnetic field, what force is exerted on it?

03:47

### Video Transcript

A 50.0 centimeter length of wire is in a 1.20 tesla magnetic field. The current in the wire is 8.00 amps. And a force of 2.40 newtons acts on the wire. What is the angle between the wire and the magnetic field? If the wire is rotated to be at an angle of 90.0 degrees to the magnetic field, what force is exerted on it?

We can call the length of wire, 50.0 centimeters, capital 𝐿. And the strength of the magnetic field, 1.20 teslas, we’ll call 𝐵. The current in the wire of 8.00 amps, we can call 𝐼. And the magnetic force magnitude of 2.40 newtons, we’ll name 𝐹 sub 𝐵.

In part one, we want to solve for the angle between the wire and the magnetic field. We’ll call that angle 𝜃. And in part two, we want to solve for a magnetic force. We’ll call that force 𝐹. We can start off solving for 𝜃 by drawing a diagram of the situation.

We have a wire of length 𝐿 that carries a current 𝐼. The wire exists in a magnetic field 𝐵 which, relative to the wire, is pointed at an angle 𝜃. To solve for this angle, we can recall a mathematical relationship for the force on a current-carrying wire. The magnetic force, 𝐹 sub 𝐵, exerted on a current-carrying wire is equal to the product of the current in the wire, 𝐼, times the length of the wire, 𝐿, times the magnitude of the magnetic field the wire is in, 𝐵, multiplied by the sine of the angle between the magnetic field and the wire direction. Considering this relationship for our scenario, we see we’ve been given all the elements of this relationship, except for 𝜃.

When we rearrange to solve for 𝜃, we see that it’s the inverse sine of 𝐹 sub 𝐵 over 𝐼𝐿𝐵. When we plug in for the terms in these parentheses, being careful to use units of meters for the length of our wire, and plug them into our calculator, we find that 𝜃 is equal to 30.0 degrees. That’s the angle between the wire and the magnetic field.

Moving on to part two, where we solve for magnetic force. We’re now told that our wire is rearranged with respect to the magnetic field 𝐵. We’re told that now our wire is arranged at an angle of 90.0 degrees with respect to 𝐵. And under these new conditions, we wanna solve for the magnetic force 𝐹. We can use the same basic equation as before to solve for 𝐹. But now when we write it out, we write 𝐹, the unknown we want to solve for, multiplied by 𝐼𝐿𝐵 times the sine of, we call it, 𝜃 sub 𝑓. That’s the final angle between the magnetic field 𝐵 and the current-carrying wire, which we’re told is 90.0 degrees. Since 𝐼, 𝐿, and 𝐵 remain unchanged and we’re given 𝜃 sub 𝑓, we’re now ready to plug in and solve for 𝐹.

With these values plugged in, once more being careful to use units of meters for our distance, we see that, because the sine of 90 degrees is equal to one, our expression simplifies to 𝐼 times 𝐿 times 𝐵. When we enter these values on our calculator, we find that 𝐹, to three significant figures, is 4.80 newtons. And that’s the magnetic force on this current-carrying wire.