A uniform conducting rod of length ℓ and mass m sits ...

Question

A uniform conducting rod of length ℓ and mass m sits atop a fulcrum, which is placed a distance ℓ/4 from the rod’s left-hand end and is immersed in a uniform magnetic field of magnitude B directed into the page (Fig. 27–54). An object whose mass M is 7.0 times greater than the rod’s mass is hung from the rod’s left-hand end. What current (direction and magnitude) should flow through the rod in order for it to be “balanced” (i.e., be at rest horizontally) on the fulcrum? (Flexible connecting wires which exert negligible force on the rod are not shown.)

Accepted Answer

Welcome back. Everyone in a physics experiment, a conducting rod of length 1.5 m and mass 2 kg is placed on a pivot 0.375 m from its left end. A mass that is six times the mass of the rod is hung from the rod's far left end. The setup is within a 0.3 tesla magnetic field pointing out of the screen, determine the current that must flow through the rod to keep it balanced. Here, we have a diagram of our setup and uh for our answer choices, it says the current must be 55 SB 98 sc 110 amperes and the D 220 pires. Now, if we're gonna determine the current that must flow, let's first make note of all the information that we have here so far, we know that the length of our rad L is 1.5 m. We know that the mass of four rod as shown in our diagrams called capital M is 2 kg. OK. We also know that a mass common M is six times the mass of the rod. So that's six M six multiplied by 2 kg or 12 kg. OK. And we know our setup is within a 0.3 tesla magnetic field. So B equals 0.3 T. So how can we use all of this information to help us figure out the value of the current? I, what do we know? Well, recall. OK, that the magnetic force that keeps the rod balanced must be equal to the strength of the magnetic field multiplied by our current multiplied by the length of the rod. So that means that the current is going to be equal to our magnetic force divided by the strength of the field, magnetic field multiplied by its length. So far, we know the values of B and L. So if we can figure out the magnetic force, then we should be able to find the current. So if we're talking about forces, then let's try to draw a free body diagram of our setup to understand exactly what forces are acting, what forces are at work here. OK. So I'm just gonna recreate our rod. OK. And for our rod, like we said earlier, we know that it's 1.5 m. OK. And for our rod again, we know that there's a mass acting to the far left. OK. Iim. And then we have our pivot and then we have the mass of the rat itself, which we know that the mass of the rat itself will be acting through the center of mass. Through the center of the rat. Now, remember we also know, OK, that our pivot is 0.35 centimeters away from the far left end, OK? And if the mass of the rod is acting through its center, then the center of a 1.5 m length rod is going to be 0.75 m from either side. OK. And if it's 0.75 m from either side, then if we compare its distance to the pivot, 0.75 minus 0.375 equals 0.375 m. OK? And if that's the case, that means since our mass, our center of mass is here, then the mass of the rod must be sorry, the weight of the rod must also be acting through the center. OK. So let me put that in red here and let's call that force FG, the weight, the force due to gravity. Now, we also know that our mass that's hanging at the far left end also has a force. OK? A weight. So let's call that weight F one. OK. And now the to keep the right balance, we can let a magnetic force FM act downward as well to the center of the rod. OK. And that is the force that we'd like to find. Now, remember our problem, our problem tells us that the rod is balanced. So since the rod is balanced, that means the sum of the moments taken about a point will be equal to zero. So let's take the sum of the moments about the pivot and then set it to zero to help us figure out our magnetic force. So taking the counter sorry, taking the counterclockwise direction as positive, taking counterclockwise as positive. Eh then we know that the sum of the moments about the pivot equals zero. OK. So now that tells us then that our counterclockwise moment, which is uh which for F one, OK. That, that is gonna be counterclockwise. So that's gonna be 0.375 m multiplied by F one which is the mass of the rod multiplied by the acceleration due to gravity G OK, minus FG which is clockwise. So it's going to be negative. So that's 0.375 m multiplied by MG. OK. It's mass multiplied by the acceleration due to gravity minus 0.375 m. OK? Multiplied by FM. The magnetic force, all of that is going to be equal to zero. OK. No, here. Notice that all of our terms have 0.375 in common. So we can cancel that out. And I remember we're solving for FM. So this means that FM is equal to zero point as well. We cancel the result actually. So let me just fix that here. So that's gonna be equal to MG minus MGM and capital M or G multiplied by M minus M Now, if we go ahead and substitute those values, OK. We take acceleration due to gravity to be 9.8 m per second squared. We know that our mass M at five and is 12 kg minus the mass of the rod 2 kg. OK? I know when we calculate here, then we get our magnetic force to be 98 newtons. No, that we have our magnetic force we can solve for the current because no, like we said, the current equals the magnetic force divided by bl. So that's 98 newtons divided by 0.3 teslas multiplied by 1.5 m which is equal to 217.78 empires or approximately 220 empires to two significant figures. Thus a current of 220 empires must flow through the rod to keep it balanced. Therefore, D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Torque

Magnetic Force on a Current-Carrying Conductor

Equilibrium

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