Can I Use My Calculator for the Upcoming Physics Exam?

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In summary: For question 1, how would you know where the Force is at all?If you want to be absolutely sure about where the force is, you need to use your thumb. If you curl your fingers based on the current or the field, your thumb will tell you where the force is. If it is negative, then you switch.
  • #1
Alt+F4
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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa00

Problem 11

So Velocity to the Right ( Thumb along velocity)
B Field Up, Fingers with B field
So ForceOut, now how would i know clock or counter?


I
 
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  • #2
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa00

Question 20

So Thumb Against AB ( Right) , B field into page

I don't see how the force could be out of the page


Ur Palm will be facing downward
 
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  • #3
Alt+F4 said:
Problem 11

So Velocity to the Right ( Thumb along velocity)
B Field Up, Fingers with B field
So ForceOut, now how would i know clock or counter?
Assume you are to view the loop exactly as shown. (You found the direction electromotive force on the charge in the moving edge of the loop. That tells you the direction of the current.)
 
  • #4
Alt+F4 said:
Question 20

So Thumb Against AB ( Right) , B field into page

I don't see how the force could be out of the page
It's not.

Ur Palm will be facing downward
Are you using your right hand?
 
  • #5
Doc Al said:
It's not.


Are you using your right hand?
wait so B field isn't into the page?
 
  • #6
Alt+F4 said:
wait so B field isn't into the page?
nevermind thanks
 
  • #7
Doc Al said:
Assume you are to view the loop exactly as shown. (You found the direction electromotive force on the charge in the moving edge of the loop. That tells you the direction of the current.)
i don't get this, sorry. Basically how i got it i was like the arrow is pointing down so it has to be clockwise. Is that good enough. How would i line up my thumb pointing down and my fingers pointing up
 
  • #8
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp05

Question 8. How is it possible that i can point my thumb down and somehow have my fingers point up. there has to be a way around this rule. thanks for anyones help
 
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  • #9
Alt+F4 said:
i don't get this, sorry. Basically how i got it i was like the arrow is pointing down so it has to be clockwise. Is that good enough. How would i line up my thumb pointing down and my fingers pointing up
For # 11. It's a bit easier to see if you imagine rotating the picture so that the z-axis is directly out of the page. B is parallel to the y-axis, and v is in the x-y plane pointing down and to the right. To rotate v into B through the smallest angle you would need a counterclockwise rotation. If you hold your right hand so that your fingers are curled counterclockwise, your thumb points in the positive z direction. One side of the loop is not moving. Two sides are moving in such a way that their contributions cancel. The remaining side is the only one you need to worry about. The force is in the direction of your thumb, out of the page.

For #20 you are correct that the field from I1 is into the page on the side of the loop. The current in a-b is to the right. Curl your right hand fingers from pointing right to pointing into the page. The force is in the direction of your thumb. This force is not out of the page.
 
  • #10
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp05

Question 8. How is it possible that i can point my thumb down and somehow have my fingers point up. there has to be a way around this rule. thanks for anyones help
It can be very difficult to twist your hand into the directions of the vectors. You don't always have to do that. For this problem you can just determine whether the smallest angle rotation from the direction of the current toward the direction of the B field is CW or CCW. Then hold your right hand so your fingers curl CW or CCW and your thumb will point either out of the page or into the page in the direction of the force.
 
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  • #11
OlderDan said:
It can be very difficult to twist your hand into the directions of the vectors. You don't always have to do that. For this problem you can just determine whether the smallest angle rotation from the direction of the current toward the direction of the B field is CW or CCW. Then hold your right hand so your fingers curl CW or CCW and your thumb will point either out of the page or into the page in the direction of the force.
hmm can you explain more? Basically in lecture and discussion they taught us the most basic stuff

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp02

So For question 1, how would you know where the Force is at all?

Another Question

If i Just Curl my fingers based on the Current or B Field, my thumb will tell me where the force is right? if it is negative then i switch. So Can i just somhow ignore Velocity?
 
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  • #12
Alt+F4 said:
hmm can you explain more? Basically in lecture and discussion they taught us the most basic stuff

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp02

So For question 1, how would you know where the Force is at all?

Another Question

If i Just Curl my fingers based on the Current or B Field, my thumb will tell me where the force is right? if it is negative then i switch. So Can i just somhow ignore Velocity?
For any moving positive charge, the right hand rule will tell you the direction of the magnetic field. Point your thumb in the direction of motion (velocity) and your findgers will curl in the direction of the magnetic field. The field circulates around the line on which the charge is moving. If a positive charge is moving toward you, the field is CCW. If it is moving away from you, the field is CW. A current is just a group of charges all moving at the same time. For a straight wire they are all moving in the same direction, so point your thumb in the direction of the current.

For the watch in #1 the center of the watch is always the same distance from the moving charge. The question does not ask about force; it asks about the direction of the field produced by the charge. There will be no force on the moving charge unless it is moving through a field produced by other moving charges. It would be hard to hold your hand with your fingers curled around the direction of motion. But you can rotate the picture either physically or in your mind to move the charge to the bottom of the figure. Then it will be easy. Convince yourself that the field in the middle of the watch is the same for any time of day.

Once you realize that the charge on the watch hand always produces the same field in the center of the watch, imagine that you have a whole ring of charges moving together. Each charge produces the same field at the watch center, so all those fields add together to make a stronger field. What you have in this case is a current loop. This leads to a different right hand rule for a current loop. From doing this problem, you should be able to come up with this rule. If you curl your fingers in the direction of the loop of current, what is the direction of the field within the loop. Actually, the field strength varies with distance from the center of the loop, but every point within the plane of a current loop has the same field direction.

The force on a moving positive charge from an existing magnetic field (created by other moving charges) has a different right hand rule. Hold your hand so that your fingers curl in the same sense as a smallest angle rotation from the v direction to the B direction. Then your thumb points in the direction of the force. Again, a current is a group of moving charges, so the direction of current is the direction of velocity. It's not that you are ignoring velocity, it is that velocity is in the same direction as the current.

You are correct that if the charges are negative, everything is reversed. You can use left hand rules for negatvie charges. If for some given configuration you reverse a velocity direction, the force is reversed. If you reverse a field direction the force is reversed. If you reverse both the velocity and the field what happens?
 
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  • #13
thanks
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp03

Question 4
Ok so let's see if i did this right

Pointed Index Finger with The arrow
So Thumb was up
B-Field Had to be into page so when i rotated my fingers to make B into page it went CounterClockwise?

Right logic at all/
 
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  • #14
Alt+F4 said:
thanks
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp03

Question 4
Ok so let's see if i did this right

Pointed Index Finger with The arrow
So Thumb was up
B-Field Had to be into page so when i rotated my fingers to make B into page it went CounterClockwise?

Right logic at all/
The rotation from v toward B in this case is from pointing to the right to pointing into the page, so your thumb points up and the force on the positive charges in the loop is upward. The current will flow CCW. That is correct.

If the B field had been out of the page, you would point your fingers to the right and then rotate them out of the page. Your thumb would then point down and the current would be CW.

If the loop had been farther to the right so it was completely contained within the region of the field what would happen?
 
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  • #15
Quick QUestion on this
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp03

Problem 15? How come current is 2.5

What i did was

Vmax = Imax Z

imax = .013645.

So Vc = Imax Xc
(.013645*176.83) = 2.4129 but for some reason they used I = 2.5 as a current. Thanks for ur help
 
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  • #16
OlderDan said:
The rotation from v toward B in this case is from pointing to the right to pointing into the page, so your thumb points up and the force on the positive charges in the loop is upward. The current will flow CCW. That is correct.

If the B field had been out of the page, you would point your fingers to the right and then rotate them out of the page. Your thumb would then point down and the current would be CW.

If the loop had been farther to the right so it was completely contained within the region of the field what would happen?
no rotation since no change in flux?
 
  • #17
Alt+F4 said:
Quick QUestion on this
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp03

Problem 15? How come current is 2.5

What i did was

Vmax = Imax Z

imax = .013645.

So Vc = Imax Xc
(.013645*176.83) = 2.4129 but for some reason they used I = 2.5 as a current. Thanks for ur help
You were given the current in the circuit. It looks like you thought you were given the source voltage. The max current is 2.5 amps
 
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  • #18
Alt+F4 said:
no rotation since no change in flux?
Right you are. From the perspective of force related to velocity and field strength, all the charges experience an upward force. The top of the loop will become somewhat more positive than the bottom, but there will be no circulation of charge around the loop, so no current will flow.
 
  • #19
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp06

Question 9

So is I Max just 5* sin 25 = 2.11
so how would i find power
 
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  • #20
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp06

Question 11

I wanted to use this forumla B = UoI/ 2 pi R but i have no idea how to use it cause i keep getting a big number
 
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  • #21
one more,

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp06

question 24.

i did this like 20 times in other exams, i don't know why it doesn't work here but

F= ILBSintheta = MG

(X)(.1)(.5) = (9.8)(1)
X = 196Amps, but the answer is 98 Amps what gives?
 
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  • #22
Last Question of the day i promise :), thanks for everyones help

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa05

Question 3

Why is it just (3)(.7*.2*2) = .084 Why not take into account the left side. Thanks
 
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  • #23
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp06

Question 9

So is I Max just 5* sin 25 = 2.11
so how would i find power
That depends on what you are supposed to know. P = IV. Since I and V are out of phase, the maximum power is not ImaxVmax. Use calculus to find the maximum product, or a graphing calculator, or what your text told you about power factor.
 
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  • #24
OlderDan said:
That depends on what you are suppose to know. P = IV. Since I and V are out of phase, the maximum power is not ImaxVmax. Use calculus to find the maximum product, or a graohing calculator, or what your text told you about power factor.
we have no Text book, so i would graph this equation I(t) = 5 sin(2π60t +Φ) and look for a max?
 
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  • #25
Alt+F4 said:
we have no next book, so i would graph this equation I(t) = 5 sin(2π60t +Φ) and look for a max?
Or you could use calculus.
 
  • #26
Hootenanny said:
Or you could use calculus.
i haven't taken Calc yet :)
 
  • #27
Alt+F4 said:
i haven't taken Calc yet :)
In that case, your only option is to graph the function. If you can download a graphing program or use an online application I would do that (much easier than by hand).
 
  • #28
Hootenanny said:
In that case, your only option is to graph the function. If you can download a graphing program or use an online application I would do that (much easier than by hand).
I can use a Ti-84 i just can't seem to get the Answer they have


I did 5Sin(2Pi60X + 25 (Phasor Angle or should a number go in there) * 120 Sin 20pi60X) = i got 561.79 instead of their answer
 
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  • #29
Alt+F4 said:
I can use a Ti-84 i just can't seem to get the Answer they haveI did 5Sin(2Pi60X + 25 (Phasor Angle or should a number go in there) * 120 Sin 20pi60X) = i got 561.79 instead of their answer
Do I read it correctly? Is that what you typed into your calc?

[tex]P = V\cdot I = 120\sin(2\pi60t)\cdot\left(5\sin(2\pi60t+\Phi)\right)[/tex]
 
  • #30
Hootenanny said:
Do I read it correctly? Is that what you typed into your calc?

[tex]P = V\cdot I = 120\sin(2\pi60t)\cdot\left(5\sin(2\pi60t+\Phi)\right)[/tex]
Exactly, except for the Phasor i put in the Angle
 
  • #31
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa02

Why is Question 11 C??
 
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  • #32
All right last question and i think i will be done
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa02

Question 5

Force = VB sin theta

but there is all this other stuff,

so i did
Force = A*V*B*R / (T) = .001 so why am i off by a factor of 10^-2. Thanks for everyones help
 
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  • #33
Alt+F4 said:
Exactly, except for the Phasor i put in the Angle
Put your calculator in radian mode and enter the phase angle in radians.
 
  • #34
OlderDan said:
Put your calculator in radian mode and enter the phase angle in radians.
Nope.i get 1.7E-9, 2.2E-8 unless i have to convert it somehow
 
  • #35
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/sp06

Question 11

I wanted to use this forumla B = UoI/ 2 pi R but i have no idea how to use it cause i keep getting a big number
That is not the formula you need. That is for finding the field produced by a current in a wire.

The induced emf is the negative of the rate of change of flux through the loop. Flux is related to field strength and area. The field is constant. The area is changing at a constant rate. The current is related to the induced emf and resistance of the loop.
 
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