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Expression for current through loop as a function of time

  1. Mar 25, 2014 #1

    kez

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    1. The problem statement, all variables and given/known data

    An N-turn circular coil of radius r with a total resistance of R is placed such that the normal to its plane is parallel to the +z axis. A uniform magnetic field varies with time according to B=B0sin([itex]\omega[/itex]t) where the amplitude B0 and angular frequency ω are constants. Find an expression for the current in the loop as a function of time.

    2. Relevant equations

    1. [itex]\Phi[/itex]B=[itex]\int[/itex]BcosA
    2. ε=d/dt [itex]\Phi[/itex]B
    3. ε=iR

    3. The attempt at a solution
    I took the integral of the magnetic field to find flux (used equation 1 from above). I multiplied by N to account for the N number of loops. Then I took the derivative of the magnetic flux (used equation 2) to determine ε. After, I just plugged in ε into the 3rd equation and solved for current... But it doesn't really make sense to plus in a value that varies with time into an equation that deals only with constant values.

    I think I missed something important here, so any help would be appreciated. I've also included an attachment of the question and the answer I got using the steps I outlined above.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 26, 2014 #2
    Your work appears to be correct. It's important to note that there's nothing saying the current or potential generated is a constant value; they can be functions of time, just as the B-field is.
     
  4. Mar 26, 2014 #3

    rude man

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    If the net field piercing the coil is B0sin(wt) then what you did is OK. But if B is an external B field = B0sin(wt) then you need to realize that the net B field thru the coil is the sum of the external B field and the self-induced B field due to the current in the coil.

    With the latter assumption the current does not track the external field as -N dø/dt / R with ø = AB0sin(wt) but builds up over time with a time constant = L/R where L and R are the inductance and resistance resp. of the coil.

    Things get even dizzier if you also want to consider the effect of the induced B on the external B field. This is the problem of mutual inductance on which e.g. transformers are based.

    EDIT: I should add that even in the steady-state the effects of self-induction must be considered. The phase of current will not be in quadrature (90 deg. phase shift) to the external B field and there is also attenuation compared to ignoring the self-induced field since the induced field 'fights' the external field via Lenz's law.

    But in all the above if R is large (R/L >> 1/w) then the effects of self-induction can be ignored, since then the induced current will be small and so will the self-induced B field.
     
    Last edited: Mar 26, 2014
  5. Mar 26, 2014 #4

    kez

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    My answer was marked wrong, so I think I'll have to go back and take self-induction into account. Thanks to the both of you for your fast responses - I appreciate it a lot!
     
  6. Mar 26, 2014 #5

    rude man

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    Well, you're the one in ten who doesn't abandon the thread after their original post!
    Hint for you: Instead of just Ri = -N d(phi)/dt, add a term L di/dt.

    You can probably ignore the transient part of the answer (assume the B field was present a long time).
     
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