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Problem 56 Hard Difficulty

A conducting bar of length $\ell$ moves to the right on two frictionless rails, as shown in Figure $P 20.30 .$ A uniform magnetic field directed into the page has a magnitude of 0.30 $\mathrm{T}$ . Assume $\ell=35 \mathrm{cm}$ and $R=9.0 \Omega .(\text { a) At what constant speed }$ should the bar move to produce an 8.5 $\mathrm{mA}$ current in the resistor? What is the direction of this induced current? (b) At what rate is energy delivered to the resistor? (c) Explain the origin of the energy being delivered to the resistor.

Answer

a. 0.73 \mathrm{m} / \mathrm{s}
b. 650 \mu J
c. \text { We must use energy to move the rod inside the magnetic field and this energy is delivered to the resistor. }

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Video Transcript

that way, we are told that we have a conducting bar with a length of 0.35 meters. That's in a uniform magnetic field of 0.3 Tesla, right? And the, uh, there's a resistor here are of 9.0 bones. So for part, it says that what constant speed should the bar move to produce a current I of 8.5 million amps or 8.5 times 10 to the minus three amps? Okay, so we'll indicate this is part A. So we can go ahead and use owns law here which says that the induce cmf for potential V is equal to the current I times the resistance are but the potential V is equal to the magnetic field times the length of the bar magnet attempt the speed will go lower case the at which the bar magnet his movie. So therefore we can solve owns law the evils I are for the speed, um, lower case be we find that this is equal to i r divided by the magnetic field times the length of the bar magnet. So be times l plugging those values into this expression before and that the speed is zero 0.73 meters per second, but it also access the direction of the current here. So as the magnetic field is in a clockwise direction and the magnitude is increasing, the direction of the current will be counterclockwise to oppose the change of flux. So we can just go ahead and type out counterclockwise here, um, and asked the direction of the current so we can say maybe counterclockwise current making box, all of this and it's their solution for part A. Okay, Part B has asked us to find the rate at which the energy is delivered to the resistance. Er so this rate is the power will indicate this is part B, and power here is equal to the square of the current times. The resistor are so if you plug in the current value in the resistor value, we find that this is equal to 6.43 Right? Excuse me. 6.56 point 46.5 times 10 to the minus four. And the power here is watts, which is energy per second box. It is their solution for part B. Okay. And lastly for part C. It asks to explain the origin of the energy that's being delivered to the resistance, so the origin of the energy that is being delivered to the resistor is going to be an external mechanical force. This is because the power that is generated by pulling the bar magnet will be equal to the electric power generated and that is utilized for heating the resistor. So the power generated by moving the bar manic magnet generates the electric power, and that's what heats up three sister. So let's go ahead and type out here. And since that movement of the bar magnet is an external mechanical force, we can say external mechanical force on far magnet when we can go ahead and box, that is your solution for part C.

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