slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Lecture 20 Discussion PowerPoint Presentation
Download Presentation
Lecture 20 Discussion

Loading in 2 Seconds...

play fullscreen
1 / 14

Lecture 20 Discussion - PowerPoint PPT Presentation


  • 71 Views
  • Uploaded on

Lecture 20 Discussion.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Lecture 20 Discussion' - axel


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Lecture 20

Discussion

slide2

[1] A rectangular coil of 150 loops forms a closed circuit with a resistance of 5 and measures 0.2 m wide by 0.1 m deep, as shown below. The circuit is placed between the poles of an electromagnet which is producing a uniform magnetic field of 40 T. The magnet is switched off, causing the magnetic field to drop to zero in 2 s. (The loops are parallel with the faces of the electromagnet.)

a) Compute the average induced potential in the circuit.

b) Determine the average current in the circuit.

c) Indicate on the drawing the direction in which the induced current flows.

slide3

C] The magnetic field from the magnet is up, but the flux is decreasing. So the magnetic field from the current induced in the loop(s) will also be up. Thus the current flow is left-to-right across the front of the loop.

slide4

[2] A conducting rod of length l moves on two (frictionless) horizontal rails, as shown to the right. A constant force of magnitude

moves the bar at a constant speed of

through a magnetic field

directed into the page. The resistor has a value

(a) What is the current through the resistor R?

(b) What is the mechanical power delivered by the

constant force?

slide5

When the conducting rod moves to the right, this serves to increase the flux as time passes , so any induced current wants to stop this change and decrease the magnetic flux. Therefore, the induced current will act in such a way to oppose the external field (i.e., the field due to the induced current will be opposite to the external field). This must be a counterclockwise current.

To find the current, we only need to find the motionally-induced voltage and then apply Ohm’s law. The rod is just a bar of length l moving at velocity v in a magnetic field B. This gives us an voltage ∆V , and Ohm’s law gives us I:

slide6

Keep in mind that v is velocity, while ∆V is voltage. Of course, the problem is now that we don’t know B. We do know that the external and magnetic forces must balance for the rod to have a constant velocity however. Constant velocity implies zero acceleration, which implies no net force. If this is to be true, the applied force must exactly balance the magnetic force on the rod. For a conductor of length l carrying a current I in a field B, we know how to calculate the magnetic force

Plug that into the first equation:

You should get I=0.5 A.

slide7

What about the power? Conservation of energy tells us that the mechanical power delivered must be the same as the power dissipated in the resistor, or

You should get 2 W.

Alternatively, you note that power delivered by a force is

where θ is the angle between the force and velocity. In this case, θ = 0, so

slide8

[3] Consider the arrangement shown in Figure. Assume that R = 6.00 Ω, L= 1.20 m, and a uniform 2.50T magnetic field is directed into the page. At what speed should the bar be moved to produce a current of 0.500 A in the resistor?

slide9

[4] Figure shows a right view of a bar that can slide without friction. The resistor is 6.00 Ω and a 2.50T magnetic field is directed perpendicularly downward, into the paper. Let L= 1.20 m.

(a) Calculate the applied force required to move the bar to the right at a constant speed of 2.00 m/s.

(b) At what rate is energy delivered to the resistor?

slide10

[5] A 6 m piece of wire that has a circular cross-section of radius 0.321 mm has a resistance of 27.6 Ω.

slide12

[6] Consider the mass spectrometer. The magnitude of the electric field between the plates of the velocity selector is 2 500 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.035 0 T. Calculate the radius of the path for a singly charged ion having a mass m =2.18 * 10-26 kg.

slide13

[7] An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT. The angular momentum of the electron about the center of the circle is 4.00 * 10-25 Js. Determine (a) the radius of the circular path and (b) the speed of the electron.

slide14

[8] A proton moves with a velocity of v = (2iˆ - 4jˆ +kˆ ) m/s in a region in which the magnetic field is B = (iˆ +2jˆ - 3kˆ ) T. What is the magnitude of the magnetic force this charge experiences?

Give it in unit vector notation,