1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inductive Heater

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data

    A 34.6 -cm-diameter coil consists of 25 turns of circular copper wire 2.25 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 3.55 × 10-3T/s. Determine

    a. the current I in the loop,

    b. and the rate dU/dt at which thermal energy is produced.

    2. Relevant equations

    magnetic flux = B*A cos Theta

    I = magnetic flux / R

    R = p (L/A) where p is the resistivity of copper

    3. The attempt at a solution

    Here is what I attempted:

    First, I found the magnetic flux by using the formula: B*A cos Theta

    multiplying the area by 25 because it's in a coil and using .173 meters for the radius of the coil

    (3.55*10^-3)(25*.173) = 1.535375*10^-2

    then to find I (current) :

    I = magnetic flux / resistance

    I calculated resistance by :

    since it's a coil, I multiplied the L by 25 and the A by 25

    R = p (L/A) = 1.68*10^-8 (25*pi*.173 / 25*pi*1.125*10^-2)^2 = 0.001148

    plugging in that into the I equation, I = 1.535375e-2 / 0.001148 = 13.374 A

    This isn't the right answer and I messed something up. I would appreciate if someone could help me with catching my mistake! I appreciate it!

    Thanks in advance!
     
  2. jcsd
  3. May 16, 2015 #2
    again. why do you say that i=magnetic flux/R ?
    ##E=-d\phi/dt##.
    so i is the change in magnetic flux per unit time/R.
    what is the area of cross section of wire?
     
  4. May 16, 2015 #3
    I meant to say that i = emf/R

    The area of the cross section is pi*r^2 which is pi(.00225)^2 where .00225 is the radius of the wire

    = 1.590431e-5

    Would I only multiply the circumference of the coil by 25?
     
  5. May 16, 2015 #4
    no. 2.25 is its diameter.
     
  6. May 16, 2015 #5
    Oh that's right! My bad.

    R = p (L/A) = 1.68*10^-8 (25*pi*.173 / pi*(0.00225/2)^2 = .11482

    then I = Emf/R = B*Area of coil / R = 3.41194*10^-2 / .11482 = 0.297155 A

    Still not getting it right. When calculating for the Emf, did I do that right? or is the fact that i's a changing magnetic field need some different calculation to get the Emf?
     
  7. May 17, 2015 #6
    what is the total length of wire? your value of L in R=p(LA) is wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inductive Heater
  1. Heater and stuff (Replies: 2)

  2. Power and heaters (Replies: 4)

  3. Piston with heater (Replies: 12)

Loading...