Question Video: Finding the Length of a Current-Carrying Wire from the Force It Experiences in a Uniform Magnetic Field | Nagwa Question Video: Finding the Length of a Current-Carrying Wire from the Force It Experiences in a Uniform Magnetic Field | Nagwa

Question Video: Finding the Length of a Current-Carrying Wire from the Force It Experiences in a Uniform Magnetic Field Physics • Third Year of Secondary School

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A 50 cm current-carrying section of wire is positioned at 90° to a 0.2 T magnetic field. It experiences a force of 0.25 N. What is the strength of the electric current in the wire?

05:09

Video Transcript

A 50-centimetre current-carrying section of wire is positioned at 90 degrees to a 0.2-tesla magnetic field. It experiences a force of 0.25 newtons. What is the strength of the electric current in the wire?

Let’s start by drawing a diagram. This can be our electric-current-carrying wire. We know that it’s 50 centimetres long. And let’s say that it carries a current upwards. We also know that it’s been placed in a magnetic field. Let’s say that the field direction is to the left.

To some extent, we can choose what directions everything is pointing in as long as we follow everything that we’ve been given in the question, which we have done by the way. We’ve been told that the current-carrying wire is positioned 90 degrees to the magnetic field. The red shows the direction of the wire. And the blue shows the direction of the magnetic field.

And guess what! They’re at 90 degrees to each other. So we have satisfied all of the conditions given to us in the question. At that point, we can choose whether the current is pointing up or down and whether the field is pointing left or right. It doesn’t matter. What does matter though is that we know the field is 0.2 tesla. We also know that the current-carrying wire experiences a force.

Now the direction of this force does matter. Well, it doesn’t really matter to answering this question particularly. But it does matter because there is a specific direction of the force. This is because the direction of the force, the direction of the current, and the direction of the magnetic field only to be orientated with respect to each other in a very specific way. We’ve already defined two of them: the current and the magnetic field.

That automatically determines which direction the third one will go. And this third thing is the force. So how do we determine which direction the force is in? We use something known as Fleming’s left-hand rule. So here is a left hand, kind of. Just pretend it is, okay?

So here’s the index finger pointing towards the left, the middle finger pointing straight up, and the thumb pointing straight out of the page. By the way, a circle with a dot in it means something is coming out of the page. That’s how we represent it.

Now these three things represent each one of the things in the diagram. The index finger represents the magnetic field 𝐵, the middle finger represents the current 𝐼, and the thumb represents the force on the wire 𝐹.

Now since we’ve determined the two directions of the field and the current, the force must be out of the page, using Fleming’s left-hand rule. So we’ve determined the direction of the force. Again, it’s not important to this question per se. But it’s good to know regardless.

So what we’ve been asked to do is to find the strength of the electric current in the wire. To find the electric current in the wire, we need to recall a relationship between the force on the wire, the current in the wire, the magnetic field which it’s in, and the length of the wire.

This happens to be the following. The force 𝐹 is equal to the magnetic field 𝐵 multiplied by the current 𝐼 and the length of the wire 𝐿. Make sure that you remember that this only applies when we have the correct orientation of the current, the magnetic field, and the force. They have to satisfy Fleming’s left-hand rule if we are to be allowed to apply the equation in green. In this case, they do, so we can apply it.

What we’re looking for is the current 𝐼. So we need to rearrange the equation a bit. We can divide both sides of the equation by 𝐵𝐿. This leads to some cancelling out on the right-hand side, leaving us with 𝐹 over 𝐵𝐿 is equal to 𝐼. And we’ve got the current isolated on the right-hand side, just as we want it.

This means that we can substitute in our values. But first, we need to have a look at the length of the wire. We’ve been told that the length is 50 centimetres. Now this is a problem. We do not want lengths in centimetres. We want it in the SI units, which is metres. So we need to convert from centimetres to metres. We can do this by remembering that the word “centi” is a prefix that means one hundredth.

In other words, a centimetre is one hundredth of a metre. So we’ve got 50 centimetres. Or in other words, we’ve got 50 times one hundredth of a metre. And that simplifies to 0.5 metres, or half a metre. So now that we’ve got the conversion done, we can substitute in our values.

The current is equal to the 0.25-newton force divided by the 0.2-tesla magnetic field times the 0.5-metre length of the wire, which means that we can plug these into our calculator. Or we can even do it in our heads. So 0.5 is the same as one-half. So we’re multiplying 0.2 by a half.

In other words, halving 0.2 gives 0.1 on the denominator. So what we’ve got is 0.25 divided by 0.1. But 0.1 is the same as one tenths. So what we’ve got is 0.25 divided by one tenths or, in other words, 0.25 times 10. And 0.25 times 10 is simply 2.5. Finally, we need to check on the units. The unit of current is amps. And so our final answer is that the strength of the electric current in the wire is 2.5 amps.

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