A thin, 50.0-cm-long metal bar with mass 750 g rests on, but is n...

Question

A thin, 50.0-cm-long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 0.450-T magnetic field, as shown in Fig. E27.37 . A battery and a 25.0-ohm resistor in series are connected to the supports. (a) What is the highest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part (a). If the resistor suddenly gets partially short-circuited, decreasing its resistance to 2.00-ohm, find the initial acceleration of the bar.

Accepted Answer

Hey, everyone. So this problem is dealing with magnetic forces. Let's see what they're giving us and what they're asking from us. So we see a student connected a voltage source and a resistor in series in two metallic supports. The student then laid a copper rod of a given mass and length horizontally on the two supports. Finally, a uniform magnetic field of a given magnitude which was directed horizontally and perpendicular to the rod was introduced. So we're asked to calculate the maximum voltage the student can apply without causing the rod to levitate. And then part two is if the student reduces the resistor by 10 X while setting the voltage to be max, which will find in part one, what would the rods initial acceleration be? So the first thing you can do is recall that the magnetic force given by a current induced in a magnetic field is F equals I L B times the sine of feta. And so here we have our current is unknown. Our force is also unknown. So we don't have that. Our length term Is 30 cm. I'm gonna rewrite that as .3 m And our magnetic field magnitude is .8 Tesla. So we don't really have enough here to solve the problem. And we don't have any terms here that can relate to this maximum voltage. So we can recall that the voltage is given by V equals I R. So now we can look at this first equation and say, okay, let's solve for I and then we can solve for voltage, we don't have the force either. So we still don't have enough. So the next thing we can do is draw a free body diagram. Let's see what's going on at this point when they say that the rod is the maximum voltage without causing the rod to levitate. So while those forces are equal to each other, so you have the magnetic force which would be equal to the weight of the rod. So we can recall that the weight is just, is just mass times gravity and we can set those equal to each other. So F B equals M G and now we can use this plugged into this first equation to solve for. So let's see what that looks like. So we have I O B times the sine of theta equals M G. And again, mass is given to us in the problem, it is 400 g or . kg. And so, and data is given to us as well. It's 90° because I said it was perpendicular. So we can solve this or I And that's going to be .4 kg times gravity. It's a constant we can recall is 9.81 m per second squared all over our length. 0.3 m times our magnetic field, 0.80 tesla times sine of 90 Which just cancels to one equals 1. And so I is equal to 16. amps. So take that We have our resistance also given to us in the problem as 30 owns and we can now solve for our voltage. So 16.35 amps times 30 and That gives us a maximum voltage of 490.5V. And so that is the answer to part one. So when we look at our potential solutions, we actually only have one that has the correct answer to part one. But let's solve out part two to make sure that we are on the right track. So in part two, we know that the resistance is reduced by times. And the voltage is this V max that we just saw for and they're asking for the rods initial acceleration. So we can use our Newton's second law again, except this time, the forces aren't going to equal zero, they're going to the sum of the forces are going to equal mass times the acceleration because we are moving and it's not the broad is not at rest. So we have R F B minus W equals M A and we can rewrite F B. We already know that that is I L B times sine of theater and then weight is vast times gravity all over M equals A. Now, the current I is not going to be the same because it is dependent on resistance. So I two is going to be V over R two where R V max is the same we just saw for it 490.5 volts and our new resistance is 30 divided by 10 or three. And so that comes out to 163. amps. And so now we do have everything we need to solve for acceleration. And so that looks like 163. Amps Times the length .3 m times the magnitude of the magnetic field, .8 Tesla Times sine of minus The mass .4 kg Times gravity 9.81 m/s squared all over Our mass again is .4 kg plug that into our calculators and we get 88.29 m per second squared. And so that is the answer to part two. So let's go back up to our choice and yes, for answer, see part two is also correct. So the answer to this problem is C that's all we have for this one. We'll see you in the next video.

Struggling with Physics?

Watch next

Magnetic Force on Current-Carrying Wire practice set