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Question about Biot-Savart Law (B-field)

  • Thread starter kenok1216
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  • #1
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Homework Statement


擷取.PNG

擷取1.PNG

I=10A,L=0.5m,a=0.3m,x=0.2m

Homework Equations


upload_2016-4-19_16-38-1.png


The Attempt at a Solution


by the Biot-Savart Law
ds cross r=-dxcosθ k(direction k)
dB= μI/4π∫-dxcosθ/r^2
B=μI/4π∫-dxcosθ/r^2 (from ?? to ??) in magnitude
question:
1) what is ?? to ??
2)how dx and r respect to dθ????


[/B]
 

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Answers and Replies

  • #2
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Look up the magnetic field from a current carrying infinitely long straight wire, an application of the BS law.
 
  • #3
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Look up the magnetic field from a current carrying infinitely long straight wire, an application of the BS law.
where is the start and end point of integration? 0 to inf?
then change cosθ and r respect to x? also the direction B-field is -k?
 
  • #4
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Are you aware of what you are doing ? What BS entails ? You write dx for an integration in the y direction, you write dB and on the next line B for the same expression on the right. You want to use the Biot Savart Law, which is fine, but somehat over the top: Ampere is already fine here!
where is the start and end point of integration? 0 to inf?
No, ##-\infty## to ##\infty##: everywhere where there is a contribution to the B field.
 
  • #5
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Are you aware of what you are doing ? What BS entails ? You write dx for an integration in the y direction, you write dB and on the next line B for the same expression on the right. You want to use the Biot Savart Law, which is fine, but somehat over the top: Ampere is already fine here!
No, ##-\infty## to ##\infty##: everywhere where there is a contribution to the B field.
OK thx
 
  • #6
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Any idea what ##\theta ## is ?
 
  • #7
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Any idea what ##\theta ## is ?
Yes after check with my lecture notes and
upload_2016-4-19_18-31-7.png
for Magnetic Field for an Infinite Long, Straight Conductor
 
  • #8
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Don't see no ##\theta## :smile:

Can you derive ##B={\mu_0I\over 2\pi a}## from BS :rolleyes: ?

But I spy an ##a## and that's not the ##a## in 'your' problem statement :nb) !
 
  • #9
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Don't see no ##\theta## :smile:

Can you derive ##B={\mu_0I\over 2\pi a}## from BS :rolleyes: ?

But I spy an ##a## and that's not the ##a## in 'your' problem statement :nb) !
upload_2016-4-19_18-57-3.png
upload_2016-4-19_18-57-22.png

upload_2016-4-19_18-57-16.png

θ1=-π/2,θ2=π/2,If the conductor is an infinitely long, straight wire
hence
upload_2016-4-19_18-59-22.png

i think a is r when i used in this question
 
  • #10
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can i ask you another question?
擷取4.PNG
5.PNG

using faraday law and lens law
e.m.f=-Nd(flux)/dt
=-200A(dB/dt)
=-200(aL)(dB/dt) how i link dB/dt with velocity dv/dt=5ms^-1???
 
  • #11
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  • #12
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can i ask you another question?
View attachment 99351 View attachment 99352
using faraday law and lens law
e.m.f=-Nd(flux)/dt
=-200A(dB/dt)
=-200(aL)(dB/dt) how i link dB/dt with velocity dv/dt=5ms^-1???
You use the result of part i.
And Flux is not ##AB## but ##\int \vec B(r)\cdot d\vec A\ \ ## !
 
  • #13
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You use the result of part i.
And Flux is not ##AB## but ##\int \vec B(r)\cdot d\vec A\ \ ## !
i feel confuse that what is the changing time in this question, since v=5ms^-1 with constant velocity,the question just asking the e.m.f at x=0.2m,but at the beginning of the figure ,it alreay in x=0.2m, so how can i cal?
 
  • #14
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i feel confuse that what is the changing time in this question, since v=5ms^-1 with constant velocity,the question just asking the e.m.f at x=0.2m,but at the beginning of the figure ,it alreay in x=0.2m, so how can i cal?
6.PNG
 
  • #15
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Do the integral for the flux. Should give you something that depends on x: ##\Phi(x)## . Then ##{d\over dt} \Phi = {d\over dx} \Phi \ {dx\over dt} = \Phi' v, \ \ ## to be evaluated at x = 0.2
 
  • #16
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Do the integral for the flux. Should give you something that depends on x: ##\Phi(x)## . Then ##{d\over dt} \Phi = {d\over dx} \Phi \ {dx\over dt} = \Phi' v, \ \ ## to be evaluated at x = 0.2
do you mean dr/dt in this case? since B=μI/2πr (dB/dr)(dr/dt)=(dr/dt)(-μI/2πr^2)=5(-μI/2π(0.2)^2) for x=0.2m
 
  • #17
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No. For the flux, r is just an integration variable. ##v## is ##{dx\over dt} = 5 ## m/s.
 
  • #18
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No. For the flux, r is just an integration variable. ##v## is ##{dx\over dt} = 5 ## m/s.
i know the concept but i do not know how to solve
e.m.f=-Nd/dx(∫B dA) (5) then how can i solve it????
 
  • #19
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Try to work out ##\int B \, dA ##

Note: e.m.f=-Nd/dt (∫B dA)
 
  • #20
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Try to work out ##\int B \, dA ##
B=
Try to work out ##\int B \, dA ##

Note: e.m.f=-Nd/dt (∫B dA)
∫B dA=∫μI/2πr^2 dA <----how A line with r dA=2πrdr?
 
  • #21
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B=

∫B dA=∫μI/2πr^2 dA <----how A line with r dA=2πrdr?
can you show me some step? i have no idea now...
 
  • #22
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A is the area of the loop over which you want to integrate ##\ \ \vec B \cdot \ d\vec A\ \ ## (because B is not the same everywhere on A).
Where do you see a ##\ \ r\, dA## ? and where do you see a ##\ \ 2\pi r \, dr## ?
 
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