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### A 0.250-m-long bar moves on parallel rails that a…

03:35
University of Kansas

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Problem 32

Consider the circuit shown in Fig. E29.31, but with the bar moving to the right with speed v. As in Exercise 29.31, the bar has length 0.360 m, $R$ = 45.0 $\Omega$, and $B =$ 0.650 T. (a) Is the induced current in the circuit clockwise or counterclockwise? (b) At an instant when the 45.0-$\Omega$ resistor is dissipating electrical energy at a rate of 0.840 J/s, what is the speed of the bar?

a. clockwise.
b. $26.3 \mathrm{m} / \mathrm{s}$

## Discussion

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## Video Transcript

So we're given the circuit that is shown in 29.31 And for part, they were asked to find the induced. Whether or not the induced current is clockwise or counterclockwise. Well, the induced current is going to be acting to resist the change in the force. So using the right hand rule where the, uh, the magnetic field's coming out of the page, the velocity of the magnets to the right, using that curl that's going to be clockwise. So it's gotta be counterclockwise to reduced resist that induced change. So for part A, we can just type out here counterclockwise, okay and then for Part B and asked us to calculate the velocity, given the information of resistance, the power of a 0.84 jewels per second, the magnetic field in the length of the rod. So a lot. The voltage is equal to the velocity times a magnetic field times the length of the rod. Therefore, velocity is equal to voltage, divided by magnetic field times the length of the rod Well, power here is equal to voltage squared, divided by resistance. Therefore, voltage is equal to the square root of power times resistance. So playing that in we have a square root of power times resistance divided by magnetic field times linked to the rod. We know all those values, so plugging them into the expression, we find that the velocity is equal to 26 0.3 meters for a second weaken box set in as our solution for part B.