 Let's solve a couple of questions on forces on current carrying conductors in magnetic fields. For the first one, we have a current of 0.2 amperes which flows towards south in a straight conductor of length 0.1 meters. When a magnetic field B perpendicular to the current is turned on in the region, the conductor feels an upward force of magnitude 0.04 newtons. What are the characteristics? So, we need to figure out the magnitude and the direction of the magnetic field. And we can use this compass for reference. Alright, pause the video and first try this question on your own. Now first let's focus on magnitude and for that we can ask ourselves, what is the force on a current carrying wire in a magnetic field? That force is given by that force, the vector, this is equal to I, this is equal to I, I into L cross B into L, L cross L cross B. And if we think about the magnitude, the magnitude would be I L B sin theta. So that would be I L B sin theta. And we can take the length vector in the direction of the current. In all the cases, we can take that to be true, the length vector can be in the direction of the current always. So that means if magnetic field is perpendicular to the current, it's also perpendicular to the length vector. So that means sin theta, that will be 1 because theta is 90 degrees. So now when you plug in the values, we get 0.2 into length which is 0.1 into the magnetic field which is what we need to figure out. And the magnitude of force that is 0.04. So let's write that also. Okay, now we can take 0.2 into 0.1 which will be 0.02 that will go to the left hand side and in the denominator and it will be 0.04 divided by 0.02. So B really comes out to be equal to 0.02 Tesla. So this is 0.02. Now when you come to direction, we can use the Fleming's left hand rule and that was that if we arrange the thumb index finger and middle finger perpendicular to each other, then the thumb gave the direction of the force if the magnetic field is given by the index finger and the direction of the current given by the middle finger. And these three are perpendicular to each other. Okay, and for this case, and for this case, we know the direction of the force. So we know this direction. We know the direction of the force, and we know the direction of the current. So we also know the direction of the current that is south. We know this direction as well. We need to figure out the direction of the magnetic field. So now if we arrange the thumb and middle finger in the direction given, whichever direction the index finger is pointing to, that is the direction of the magnetic field. So the current is in the same direction as the displacement vector and all of them, they're always in the same direction and that is to the south. So pointing the middle finger to the south, that is the downwards direction, this direction. And then pointing the thumb upwards. So really the thumb should be pointing towards you and if you're doing this at home or at school or wherever, your middle finger is pointing down and your thumb should be pointing really towards you. And I'm sure you will realize that I'm also doing it live, that the index finger now is pointing to the east, it's pointing to the right. And our hand could look somewhat like this. So this is the current, this is the current that is south and the force is pointing up, this is the thumb. The force is pointing up. So the magnetic field, as a result of that, the index finger pointing to the east. And you could rest your hand on a table just like the image on the screen. In that case, you see the thumb is really pointing up towards the ceiling or the roof. But if you lift your hand and if you try to arrange the fingers in the direction given, then the thumb might be pointing towards you and that's okay. Your index finger is still pointing to the right. So that's in the east direction. This will be east. Okay, let's look at one more question. Okay, here we have a 1 meter rod carrying 3 amperes current, which is placed in a uniform magnetic field and the magnetic field has a magnitude of 5 milli Tesla and the angle this angle theta is 30 degrees. Find the force acting on the rod and we need to figure out the magnitude and the direction of the force. This arrow represents the current direction, which is going in this direction. All right, now again, just like in the previous question, let's start by writing what force is, force on a current carrying conductor. So that force, that force really is given by I L cross P and I'm now writing everything with the same color. So I L cross B. To figure out the magnitude, we can just open the cross product and that will be I L B sine theta. Now here we know what I is, that is 3, so we know this. We know length, this is a 1 meter rod. We know the magnetic field, right, 5 milli Tesla. There is also an angle given. Now this angle theta, this angle theta is always the angle between the length vector and the magnetic field vector. You can take the length vector in the same direction as the current. So this could be your length vector and the angle between the length vector and the magnetic field is really this angle and this is 90 minus theta. This is 90 minus theta. Theta is 30, so 90 minus 30, that is 60. So sine 60 degrees. This will be then 3 into 1 into 5 milli Tesla. So 5 into 10 to the power minus 3 into root 3 by 2. And this is really, this is really 30 into 10 to the power minus 3 Newtons. Now we need to report the answer in milli Newtons. We should have seen that first. We then wouldn't have to convert milli Tesla to Tesla, right? So okay, this is then really just 13 milli Newtons because 1 milli Newton is equal to 10 to the power minus 3 Newtons. And now let's think about the direction. So we need to figure out the direction of the force. Again, we can use the Fleming's left hand rule, but here the rod is angled, right? It's angled with the magnetic field. And whenever there are such cases, we can try and resolve the length component, right? So there will be, the length vector really is at an angle. There will be a vertical component. There will be a horizontal component. The horizontal component will really not give rise to any sort of force because that could be either in the same direction as the magnetic field, just like in the case that you see, or in the opposite direction, like the angle could be zero or 180. In either case, sine of that will be zero. So only the component that is perpendicular to the magnetic field, only that will give rise to some force. And that will be this component in this case because the length vector is taken to be in this direction. So if we do resolve it, there will be a vertical component like this. So the horizontal component not playing any role. Let's ignore that. Let's only look at the vertical component. So here, if we arrange, again, our left hand in the directions that are given to us. So we are taking this vector, considering this length vector, the only the vertical component. So FBI, force is given by thumb, magnetic field is index finger and current is the middle finger. Now, if we use the Fleming's left hand rule, we can arrange the hand like this. This current is really going away from us. We can think of it as going away from us. We have the magnetic field lines which are to the right. So this is your index finger B, this is your current I, and the force, the direction of the thumb, in this case, it's in the downward direction. So that is what the direction is. Now, we could imagine this rod lying in front of us on the table. And like if you do that, imagine you're looking at this rod from the top. So if you do that, then if you arrange your index finger, which is pointing to the right, and your middle finger, which will really be pointing away from you, further away from you. So then you will realize that the thumb is pointing down, pointing inside the table or inside the desk. And that really means into the plane. So the direction of the force for this one is into the plane. All right, you can try more questions from this exercise in the lesson. And if you're watching on YouTube, do check out the exercise link which is added in the description.