Calculate the induced EMF for angle α

  • #1
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I have a question in my book and it’s confusing me a bit. I tried to search online for similar solved problems but couldn’t succeed. So here it goes:


Calculate the induced EMF in a conductor loop when the angle between [tex] \vec{A} ~ and~ \vec{B} ~is~ changed ~from ~{0 °}~ to ~{α °} [/tex] in 1 second: [tex] Δt = 1s[/tex]


So I solved it like this: [tex]ε = A \cdot B \cdot {cos(\alpha) - cos(0) \over Δt}[/tex]


Where: [tex] B = 2.8 \cdot 10^{-2} T[/tex] and diameter of the loop is: [tex]D = 5.4 cm [/tex]


So using the above method, I should get and induced Voltage of [tex] -9 \cdot 10^{-6} V[/tex], if I rotate the loop from an initial 0 ° Orientation to 30 °. Would this be wrong? If so, could you explain that to me?
 

Answers and Replies

  • #2
BvU
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Hello Yousuf, :welcome:

Why would it be wrong ? You use the correct formula and do the math correctly, so what's the problem ?

There is one remark to be made: it is not very neat to speak of 'the' EMF when in fact this is an average EMF (from ##\ B\Delta A/\Delta t\ ## ) and one second can be a long time in electronics.
 
  • #3
rude man
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I have a question in my book and it’s confusing me a bit. I tried to search online for similar solved problems but couldn’t succeed. So here it goes:
Calculate the induced EMF in a conductor loop when the angle between [tex] \vec{A} ~ and~ \vec{B} ~is~ changed ~from ~{0 °}~ to ~{α °} [/tex] in 1 second: [tex] Δt = 1s[/tex]
How do you define this angle? The rigorous way is the angle between B and the normal to the plane of A. But many think of the area vector as being in the plane which is not good since that vector can have any orientation within the A plane.

And your answer depends on that definition.
 

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