EMF induced in solenoid by current in a loop.

  • Thread starter bobred
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  • #1
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Homework Statement


A square loop with side-length a is positioned at the centre of a long thin solenoid, which has radius r (with r>a), length l and N turns. The plane of the loop is perpendicular to the axis of the solenoid, find [itex]V_{emf}[/itex] induced in the solenoid

Homework Equations


[itex]M=\frac{\Phi}{I}[/itex]

[itex]\Phi=\mathbf{B}\cdot\mathbf{A}[/itex]

[itex]V_{emf}=-M\dfrac{\mathbf{\textrm{d}}I}{\textrm{d}t}[/itex]

The Attempt at a Solution


The magnetic field created by the current flowing through the loop is complex and the flux varies throughout the solenoid. By using the fact that the mutual inductance is the same for the coil and the solenoid we can find the emf generated in the solenoid.
If we take the magnetic field produced by a solenoid as

[itex]\mathbf{B}=\mu_{0}nI_{sol}\:\mathbf{e}_{z}[/itex] where [itex]n=\dfrac{N}{L}[/itex]

Then the flux through the square loop is

[itex]\Phi=\mathbf{B}\cdot\mathbf{A}=\mu_{0}nI_{sol} \times a^{2}=\mu_{0}na^{2}I_{sol}[/itex]

Now the mutual inductance is

[itex]M=\dfrac{\Phi}{I_{sol}}=\mu_{0}na^{2}[/itex]

The induced emf is [itex]I_{loop}=I_{0}\sin\omega t[/itex]

[itex]V_{emf}=-M\dfrac{\mathbf{\textrm{d}}I_{loop}}{\textrm{d}t}=-\mu_{0}na^{2}\omega I_{0}\cos\omega t[/itex]

Is this correct?
 

Answers and Replies

  • #2
30
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I think you are leaving out some things in the problem statement. What is the current through the square loop?
 
  • #3
173
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Hi, yeah the current in the loop should be [tex]I_{loop}=I_{0}\sin\omega t[/tex].
Sorry, ignore the expression after 'The induced emf is'
 

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