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IElectromagnetism - Inductince in two small wire loops

  1. Mar 16, 2009 #1
    I'm only stuck at the very end of this quesstion but ill wirte out the whole thing anyway.

    Two tiny wire loops, with (vector) areas [itex]\vec{A_1},\vec{A_2}[/itex] respectively are seperated by a p[osition vector [itex]\vec{r}[/itex] with [itex]r^2>>A_1,A_2[/itex]

    a) Treating the loops as magnetic dipoles, find their mutual inductance. Confirm that it is symmetric under exchange of loops.

    b) A current I1 is flowing in loops 1 and a current I2 is then slwoly built up in loop two. How much work must be done against the mutually induced emf to keep the current I2 flowing? How does this result compare with that in magnetostatics for the work requiredto bring up a dipole of moment [itex]\vec{m}[/itex] from infinity in a magnetic field [itex]\vec{B}[/itex]

    I have succesfully done a) and checked it with my tutor.

    The mutual inductance was [itex]M=\frac{\mu_0}{4 \pi} \frac{\left[3(\vec{A_1} \cdot \vec{r})(\vec{A_2} \cdot \vec{r}) - \vec{A_1} \cdot \vec{A_2} r^2\right]}{r^5}[/itex]

    clearly this is symmetric under relabbeling of [itex]1 \rightarrow 2[/itex] as required and so we have just said [itex]M_{12}=M_{21}=M[/itex]

    As for part b) I have found the work that must be done as follows:

    [itex]\varepsilon_1=-\frac{d \Phi_1}{dt}[/itex] but [itex]\Phi_1=I_2 M[/itex]

    and so [itex]\varepsilon_1=-M \frac{d I_2}{dt}[/itex]

    we know that [itex]\frac{dW}{dt}=- \varepsilon_1 I_1=M I_1 \frac{d I_2}{dt} \Rightarrow W=MI_1I_2[/itex]

    AS FOR THE PART about bringing a dipole from infinity. so far I have:

    [itex]\vec{F}=\nabla (\vec{m} \cdot \vec{B})[/itex] and so

    [itex]W=\int_{\infty}^{\vec{r}} \vec{F} \cdot \vec{dr} = \vec{m} \cdot \vec{B}[/itex]

    the field from a dipole is [itex]\vec{B}=\frac{\mu_0}{4 \pi} \frac{3(\vec{m} \cdot \vec{r})-\vec{m} r^2}{r^5}[/itex]
    but substituting that in should get me back to [itex]W=MI_1I_2[/itex] should it not? i must be doing something wrong here.
    would it be ok to say that the arbitrary field [itex]\vec{B}[/itex] is generated by a dipole of moment [itex]\vec{m_1}=I_1 \vec{A_1}[/itex] and that the dipole referred to in the question has moment [itex]\vec{m_2}=I_2 \vec{A_2}[/itex] as then this does match up with the answer from the first part.

    also am i correct in saying [itex]\varepsilon_1[/itex] is aback emf?
     
    Last edited: Mar 16, 2009
  2. jcsd
  3. Mar 16, 2009 #2

    gabbagabbahey

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    Re: Electromagnetism

    Are you sure about that? Be careful with you negative signs; the work done by you=-work done by the field.
    Also, you left out an r in your expression for B...i fixed it in the above quote.

    Yes, the question tells you to treat the current loops as being far enough away from echother, and small enough in diameter as to assume that they are pure dipoles [tex]\vec{m}_i=I_i\vec{A}_i[/tex]

    But when you correctly do the calculation for the work required to bring the dipole in from infinity, you should find that your two results differ by a negative sign....why?:wink:
     
  4. Mar 16, 2009 #3
    Re: Electormagnetism

    the second one is work done against the field and the first one is work done by the field at a guess - i'm not sure.

    how do i get the negative in the second equation, should i have

    [itex]W=-\int_{\infty}^{\vec{r}} \vec{F} \cdot \vec{dr}[/itex]???
     
  5. Mar 16, 2009 #4

    gabbagabbahey

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    Re: Electormagnetism

    That'll work, yes.

    No, the reason for the sign difference is very different. Both methods calculate the work done by you; but the magnetostatic calculation neglects the energy required to maintain the __?__ in the loop. It also suggests that work is done by the magnetic field, which you should know, is never the case.

    Your first calculation produces the correct energy, the magnetostatic one does not. For an in depth discussion of why the two calculations curiously turn out to diiffer only by a sign change, see R.H. Young, Am. J. Phys 66, 1043 (1998) --- In which Young answers Griffiths' question on why the Hamiltonian is [itex]-\vec{\mu}\cdot\vec{B}[/itex]--- and references cited therein.

    You may also find Sam Park's and clem's posts in this thread useful.
     
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