Maple Quiz 18.3

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Maple Quiz 18.3

Post  maple sy on Wed Feb 18, 2009 2:34 am

did anyone get #1 and 4? i used the equation in the lecture notes but im not getting it right.

maple sy
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Re: Maple Quiz 18.3

Post  entee on Wed Feb 18, 2009 9:09 pm

#1 is simply I(current) * A(area)

make sure you convert the length to meters. & just square it to find the area. the answer is POSITIVE.

I can't get #4 either...anyone know?

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Re: Maple Quiz 18.3

Post  entee on Wed Feb 18, 2009 9:19 pm

actually i figured out #4...it's because you have to convert your final answer to electron volts
1 eV = 1.602 J

answer is NEGATIVE

entee
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Re: Maple Quiz 18.3

Post  juggy on Thu Feb 19, 2009 5:35 pm

1) Area *current
2) PE=((area*0.0001)*I)*(magnetic field)*cos0 … answer is negative.
3) u=(qvr)/2 … q=1.6*10^-19 … answer is negative.
4) This one is really weird… but… multiply the two given number together. Multiply that by two. Then divide it by 1.602. and then finally, multiply it by 10^19.

juggy

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Re: Maple Quiz 18.3

Post  physics on Thu Feb 19, 2009 5:37 pm

...

physics
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how

Post  thanks on Fri Feb 20, 2009 3:28 pm

thanks very much for all the help guys.. this site has been very helpful!
but physics has been very difficult subject for me.. how do you study this course? are there any advice that you would like to share?

thanks
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Re: Maple Quiz 18.3

Post  me on Fri Feb 20, 2009 5:09 pm

#4 still wont work for me...i tried doing it following the example on sheet 38 and then when i got the answer in joules i divided it by 1.602 to put into eV but it still didnt work..any other suggestions?

me
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ques 4

Post  me on Fri Feb 20, 2009 5:52 pm

okay nvm i got it to work but why do you multiply it by 10^19?

me
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#2

Post  sck105 on Sat Feb 21, 2009 6:12 pm

I have tried number 2 so many times but its not working for me can someone please post a detailed description of how to do this problem?

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q2

Post  hwilson on Sat Feb 21, 2009 6:19 pm

I need help with number 2 as well can someone plz explain?

hwilson
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question 2

Post  craziech on Sat Feb 21, 2009 6:37 pm

I cant figure out #2 either any ideas?

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#2

Post  WHATEVER on Sat Feb 21, 2009 6:54 pm

DON'T USE COSINE 0. YOU HAVE TO USE THE ANGLE THEY GIVE YOU IN THE PROBLEM!

WHATEVER
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question 2

Post  :) on Sat Feb 21, 2009 7:00 pm

thank you guys soo much for your help!

so far i am still stuck on number 2 however.

I thought it was just -((Area x 1E-4) * (Current I) * (Magnetic Field B) * cos(90)

and the answer would be negative

but this answer is not accepted o.o

is there something else i am leaving out or missing?

thank you so much for your time and help!

:)
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question 2

Post  :) on Sun Feb 22, 2009 12:35 am

okay icic. instead of the cos 90, i had to use the angle given, duh! lol

thank youu!! Very Happy

:)
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#2

Post  lost on Sun Feb 22, 2009 10:27 am

No matter what I do I still cannot get #2.

Here are my numbers: area=31.15cm^2, current=.1609A, angle relative to direction of uniiform magnetic field=166.84, magnetic field=0.15T

When I put it in to the equation:

PE= ((31.15*.0001)*.1609)*(0.15)*cos(166.84)

I've tried this so many times and it keeps telling me I'm wrong!
Can anyone help?

lost
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lost #2

Post  super mo on Sun Feb 22, 2009 11:41 am

I re-did your problem and got an answer of 0.000073. I did notice that you did not use a "-" sign in front of the formula that you posted. I also put in ...*(cos(16.84)) when I did the calculaiton. This might have gummed up the answer as well. Let me know how you made out.

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Re: #2

Post  lost on Sun Feb 22, 2009 1:21 pm

Thanks so much super mo...it worked!

lost
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lost

Post  super mo on Sun Feb 22, 2009 2:39 pm

cool!!! Very Happy

super mo
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#4

Post  #4 on Sun Feb 22, 2009 3:33 pm

plz any help with #4

#4
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Q #2

Post  phent on Sun Feb 22, 2009 3:43 pm

A wire loop with a 26.91 cm2 area carrying a 0.2981 A current has its magnetic moment directed at an angle of 151.380 relative to the direction of a uniform magnetic field of 0.15 T in which the loop is placed. What is the potential energy of the magnetic dipole (see sheet 32-38) ? Indicate with a positive (negative) sign whether the minimum potential energy is at 900 (00).

(( A* .0001)*I)*(Magnetic field)*(cos151.380)

I have tried many times this question and when I do my calculation i get the asnwer of -1.056E-8. I don't know why its not working.

Any help?

phent
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4

Post  helpp on Sun Feb 22, 2009 4:01 pm

guys I can't get 4

helpp
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#4

Post  super mo on Sun Feb 22, 2009 4:20 pm

Look at pg 38. On the bottom it states: delta U= 2uB Just plug in what they gave you and divide by1.6*10-19. You need to divide by this number in order for your answer to be in electron volts. ALWAYS scan through previoius posts before asking for help. You can usually find the answer you are lookng for.

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Phet

Post  super mo on Sun Feb 22, 2009 4:26 pm

I plugged in your #'s and did not get that answer look at what I posted agove and see if that helps EX: Your formla did not contain a "-" sign

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#2

Post  JJ on Mon Feb 23, 2009 3:12 pm

your calculator has to be in degrees for number 2..... i felt really silly when i got found that out

JJ
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Re: Maple Quiz 18.3

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