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Homework Help: Clarification on induced current/Bfields for solenoids

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm working on induced currents/b-fields from Lenz's Laws and two different problems use the same equation but one has an extra variable and I don't know when to add this equation.

    Problem 1:
    In a physics laboratory experiment, a coil with 180 turns enclosing an area of 11.6cm^2 is rotated during the time interval 3.10×10−2s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00×10−5T. What is the total magnitude of the magnetic flux ( Phi_initial) through the coil before it is rotated?

    Problem 1 solution equation:
    Phi = BAcostheta
    This relation is just for one loop, and when we are calculating the effect of all the loops in the coil we must remember to multiply by N, the total number of loops. In other words Phi_initial = N*Phi_1.

    ^Now this completely makes sense to me, but I just did another problem:

    Problem 2:
    A 100-turn 8 cm diameter coil is made of 0.5 mm diameter copper wire. A magnetic field is perpendicular to the coil. At what rate must B increase to induce a 2 A current in the coil?

    I did this problem perfectly, but I I'm off by a factor of 100 because I divide my flux by 100 to account for the numbers of turns.

    My question is: why do I need to account for the turns in one problem and not another? ie. In the future, how will I know what to do?

    Thank you!
  2. jcsd
  3. Apr 28, 2013 #2
    When you divide the flux that you have obtained by 100, you are finding the flux liked with each turn. But the net emf induced will be because of the net change in flux. i.e. the sum of flux change linked with each of the 100 coils.
  4. Apr 28, 2013 #3

    rude man

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    Homework Helper
    Gold Member

    No. the flux phi is stll BA. It's only the emf that will get multiplied by 100: emf = -N d(phi)/dt.
    We have
    emf = -N d(phi)/dt and then i = emf/R so you need to know the resistance of the coil also. You can compute R from the wire length, diameter and conductivity of Cu.

    Phi = B*A, A = area, so emf = -NA dB/dt. Solve for dB/dt.

    The solutions of problems 1 and 2 are perfectly consistent.
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