Inductive Heater

1. May 16, 2015

Angie K.

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!

2. May 16, 2015

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?

3. May 16, 2015

Angie K.

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?

4. May 16, 2015

no. 2.25 is its diameter.

5. May 16, 2015

Angie K.

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?

6. May 17, 2015