Magnetic Force,EMF, Flux question

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The discussion focuses on three questions related to magnetic force, magnetic flux, and induced electromotive force (EMF) in a coil. For the magnetic force on a loop with a given magnetic field and current, the calculation of 0.8 N is confirmed, but clarification on the direction is needed based on the right-hand rule. The magnetic flux calculation initially included an incorrect factor of pi, as the loop is square, not circular, and the correct approach involves squaring the length of one side. The induced EMF generated when the loop is pulled from the magnetic field is calculated correctly at 0.032 V, with the resulting current through a 4-ohm resistor being 0.008 A. Overall, the calculations are mostly accurate, with minor adjustments needed for the flux and force direction.
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I have three questions that I think I have the answers to, but would like someone to tell me if I did them correctly. Thanks!
Q1. Given B= 0.8T, a coil with 1 loop with length of 20cm, a current of 5.0A, what are the size and direction of the magnetic force on the loop?
A1. Would I just use Fm= IxBxL = (5A)(.8T)(.2m) = .8 N ?
As for the direction I know it involves the right hand rule, and the current moves to the right, so would the magnetic force be pointing up ?

Q2. What is the magnetic flux through the loop?
A2. Would I use: Flux= ABcos0 = pi(.2)^2 (.8T) (cos0) = .101 Wb ??

Q3. If the magnetic field has a cut off at the top of the loop, and the loop is pulled completely out of the field upwards at v = .2 m/s:
what EMF would be generated and if the resistance was 4 ohms, what would the current be?
A3: EMF= Blv = (.8T)(.2m)(.2m/s) = .032V
O = EMF/R = .032V / 4 ohms = .008Ado these look right? thanks.
 
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nophun6 said:
I have three questions that I think I have the answers to, but would like someone to tell me if I did them correctly. Thanks!
Q1. Given B= 0.8T, a coil with 1 loop with length of 20cm, a current of 5.0A, what are the size and direction of the magnetic force on the loop?
A1. Would I just use Fm= IxBxL = (5A)(.8T)(.2m) = .8 N ?
As for the direction I know it involves the right hand rule, and the current moves to the right, so would the magnetic force be pointing up ?

Q2. What is the magnetic flux through the loop?
A2. Would I use: Flux= ABcos0 = pi(.2)^2 (.8T) (cos0) = .101 Wb ??

Q3. If the magnetic field has a cut off at the top of the loop, and the loop is pulled completely out of the field upwards at v = .2 m/s:
what EMF would be generated and if the resistance was 4 ohms, what would the current be?
A3: EMF= Blv = (.8T)(.2m)(.2m/s) = .032V
O = EMF/R = .032V / 4 ohms = .008A


do these look right? thanks.

You have to tell us in what direction the B field is pointing! (with respect to the normal of the loop, let's say)
 
nrqed said:
About Q2: It looks like you used the length of the loop for the radius!

Pat


Oh wait, it's a *square* loop (when I saw your solution I thought you had in mind a circular loop!). Then it's ok to square the length of one side, but there shoul dbe no factor of \pi there!
 
whoops, two things: i forgot to mention that in question 1 I nedd to find the magnetic force and direction on just the Lower section of the loop, and the second thing is i guess i wasnt thinking about the pi, thanks for pointing that mistake out.
If i am looking at just the lower section of the loop would it be: F= IBLSin(theta) where theta is 90??
thanks.
 
nophun6 said:
whoops, two things: i forgot to mention that in question 1 I nedd to find the magnetic force and direction on just the Lower section of the loop, and the second thing is i guess i wasnt thinking about the pi, thanks for pointing that mistake out.
If i am looking at just the lower section of the loop would it be: F= IBLSin(theta) where theta is 90??
thanks.

Ah.. ok. Yes, then theta is 90. I assume you know how to get the direction of the force...

And your answer for the induced emf seems correct.

Pat
 
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