This discussion focuses on the application of the right-hand rule in electromagnetism, specifically in determining the direction of force, magnetic fields, and current in various physics problems. Participants analyze specific questions from a physics exam, including the direction of forces in magnetic fields and the behavior of currents in loops. Key concepts include the right-hand rule for both moving charges and current loops, as well as the relationship between velocity, magnetic field, and force. The discussion emphasizes the importance of visualizing vector directions and the correct application of the right-hand rule to solve problems accurately.
PREREQUISITESStudents preparing for physics exams, educators teaching electromagnetism, and anyone interested in mastering the concepts of magnetic fields and forces in physics.
Put your calculator in radian mode and enter the phase angle in radians.Alt+F4 said:Exactly, except for the Phasor i put in the Angle
Nope.i get 1.7E-9, 2.2E-8 unless i have to convert it somehowOlderDan said:Put your calculator in radian mode and enter the phase angle in radians.
That is not the formula you need. That is for finding the field produced by a current in a wire.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 formula B = UoI/ 2 pi R but i have no idea how to use it cause i keep getting a big number
o yaaaaa thanks a lotOlderDan said: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.
all right so why am i only counting current for one side icne 196/2 = 98Alt+F4 said: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?
You are doing something wrong in your calculator. If the current and voltge were in phase you would have a maximum power of 120V*5A=600W. The phase difference reduces this somewhat, but not by a huge amount.Alt+F4 said:Nope.i get 1.7E-9, 2.2E-8 unless i have to convert it somehow
hmm well i think there is something wrong with their calculation, i have no idea how they got 25 Degrees to .436 radians, i got .4663 radians so maybe that is why i can't get their answerOlderDan said:You are doing something wrong in your calculator. If the current and voltge were in phase you would have a maximum power of 120V*5A=600W. The phase difference reduces this somewhat, but not by a huge amount.
25/180*π = .43633 not .4663Alt+F4 said:hmm well i think there is something wrong with their calculation, i have no idea how they got 25 Degrees to .436 radians, i got .4663 radians so maybe that is why i can't get their answer
ya i can't get the answer at all, i tried on diffrent calculators. i am postiive I am typing it correctlyOlderDan said:25/180*π = .43633 not .4663
Are you using the calculator to find the maximum of the product IV? How are you doing this?Alt+F4 said:ya i can't get the answer at all, i tried on diffrent calculators. i am postiive I am typing it correctly
is there an explanation for this one or is this one of the thing you memorizeAlt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa02
Why is Question 11 C??
When I do that I get 571.9W using their value of 0.436 for phase in radians, which is correct. What do you get?Alt+F4 said:i went to Y = and type in that whole equation, then hit Graph, then did 2nd Graph and found the maximum value
It is not sin180. It is sin180t with t constantly changing. You need to find the maximum rate of change of the flux and use the number of turns to find the maximum emf.Alt+F4 said:Question 16 please :)
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa05
So i tried doing it on Calculator but since sin180 = 0 i get 0
Did you find the current in the circuit? Did you compute the voltages for each of the elements from that current?Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa02
Why is Question 11 C??
This is from the left side. It is the only side in the magnetic field. The right side contibutes nothing.Alt+F4 said: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
same but the answer is 543 that is why I am so confusedOlderDan said:When I do that I get 571.9W using their value of 0.436 for phase in radians, which is correct. What do you get?
is this the same idea on how one side isn't really contibuitng?Alt+F4 said: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?
I thought you had already resolved this one (#24). For this one you only used one side of the loop when you should have used both sides. Both sides contribute to the torque, so only half the current you calculated is needed.Alt+F4 said:is this the same idea on how one side isn't really contibuitng?
silly me, thanks a lot for your help like seriously. My TA's arent even that helpfulOlderDan said:I thought you had already resolved this one (#24). For this one you only used one side of the loop when you should have used both sides. Both sides contribute to the torqe, so only half the current you calculated is needed.
hmm all right so i graphed this, Does it matter if calc is in radian or degree?Alt+F4 said:Question 16 please :)
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa05
So i tried doing it on Calculator but since sin180 = 0 i get 0
You are supposed to be finding the value of C that will make the resonant frequency of the circuit the desired broadcast frequency. How does C affect the net capacitance of the circuit?Alt+F4 said:All right last question until Nov 23rd i think
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/fa03
Question 21
So W = 2piF = (2*3.14*580*10^3) = 3.6E6
>and 1 / ( sqrt 1*10^-3 * .06*10^-6) = 1.2909 E5
what am i doing wrong
Currents parallel to magnetic fields experience no force. Currents at right angles do.Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam2/su06
Question 20
Finger Curl with current
B Field points Down
I woulda picked Rotates it so that the right side comes out of the page and the left goes in.
Yes. Always use radians for these time dependent problems. If they ask for phase in degrees, convert it from radians. If they give phase in degrees, convert it to radians. Angular frequency is always in radians/sec.Alt+F4 said:hmm all right so i graphed this, Does it matter if calc is in radian or degree?
all right but for some reason when i graph in Radian mode i seriously get like -6 and -5 values unless i am doing something wrong with itOlderDan said:Yes. Always use radians for these time dependent problems. If they ask for phase in degrees, convert it from radians. If they give phase in degrees, convert it to radians. Angular frequency is always in radians/sec.