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A 1.41 $\mathrm{m}$ bar moves through a uniform, 1.20 $\mathrm{T}$ magnetic field with a speed of 2.50 $\mathrm{m} / \mathrm{s}$ (Figure $21.60 ) .$ In each case, find the emf induced between the ends of this bar and identify which, if any, end $(a$ or $b)$ is at the higher potential. The bar moves in the direction of $(a)$ the $+x$ -axis; $($ b) the $-y$ -axis; (c) the $+z$ -axis. (d) How should this bar move so that the emf across its ends has the greatest possible value with $b$ at a higher potential than $a,$ and what is this maximum emf?

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a. the emf induced is $[2.54 \mathrm{V}]$b. the emf induced is $[3.38 \mathrm{V}]$c. the emf is zero.d. the maximum emf is $[4.23 \mathrm{V}] .$ To attain this emf, the rod must be moved upwards, that is towards the second quadrant.

Physics 102 Electricity and Magnetism

Chapter 21

Electromagnetic Induction

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Inductance

Alternating Current

Cornell University

University of Michigan - Ann Arbor

University of Washington

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Lectures

03:27

Electromagnetic induction is the production of an electromotive force (emf) across a conductor due to its dynamic interaction with a magnetic field. Michael Faraday is generally credited with the discovery of electromagnetic induction in 1831.

08:42

In physics, a magnetic field is a vector field that describes the magnetic influence of electric currents and magnetic materials. The magnetic field at any given point is specified by both a direction and a magnitude (or strength); as such it is a vector field. The term is used for two distinct but closely related fields denoted by the symbols B and H, where H is measured in units of amperes per meter (usually in the cgs system of units) and B is measured in teslas (SI units).

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A $1.41-\mathrm{m}$ bar mo…

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A metal bar is pulled thro…

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A 1.50 -m-long metal bar i…

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In Figure 19.7 , take $B=3…

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The apparatus in Figure 19…

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A uniform, 458 -g metal ba…

01:40

A metal bar of length 25 c…

01:45

A copper bar $30 \mathrm{~…

So here the induced EMF is going to be equal to the velocity times the magnitude of the magnetic field times the length. Ah, however, for this case, the induced the IMF will be equal to Vbl time. Sign of theta. Um, again Here the velocity of the bar is not perfectly perpendicular to the bar. It actually produces a magnetic force that is not aligned with the bar. So that is why we have to include the angle part a weaken say that the IMF is going to be the be our sign of Fada again. Here for part A. The velocity is in the positive ex direction. So if it's in the positive ex direction, the IMF is going to be equal to 2.50 meters per second times one point 20 Tesla's times 1.41 meters and then times sign of 37 degrees. And this is going to be equal to 2.55 volts with a ah higher potential at S O higher potential at point. Hey! Okay, again, 2.55 votes for part B. We know that if he is going to be the is in negative y direction. So here Seita, the angle between the access of the bar and the philosophy here is going to be not 37 degrees, but rather 53 degrees. So we can say that the IMF is going to be equal. Tio 2.5 times 1.2 of 2.5 meters per second time's one point to Tesla's times again, 1.41 meters on then times sign of 53 degrees. This is equaling 3.38 volts And again we have a higher potential at a now for part Z for fartsy vee is in the positive zy directions. So at this point, um, EMF is going to be equal to zero volts because the angle between the axis of the bar and the velocity is going to be zero degrees. So the velocity is parallel to the magnetic fields. So you can say that thie is parallel to the magnetic field. So tha tha tha is going to equal zero degrees and we know that sign of zero degrees equal zero, which means that the IMF well legal Zahra bolts for part D ah, in order for the in order for the potential at B To be higher, the bar must move perpendicular to its length. Uh, and we can say for which MM is 4.23 fold. So that would be the maximum EMF in max. And here for the electric potential at point B to be greater than the electric potential at point A, the bar must move upward. Ah, and to the left, towards the second quadrant again, perpendicular to its length, so perpendicular to its length will simply make the IMF the maximum in math. That's possible. And then, in order for the electric potential, it be to be greater than the electric potential a day it must move upward and to the left. So towards the second quadrant, that is the end of the solution. Thank you for watching

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