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Magnetic Flux Through a Coil

  • Thread starter ttiger2k7
  • Start date
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1. Homework Statement
A long straight wire on the z-axis carries a current of 3.0 A in the positive direction. A circular loop in the xy-plane, of radius 10 cm, carries a 5.0-A current, as shown. Point P, at the center of the loop, is 25 cm from the z-axis.

28.6a.jpg


A circular coil of four turns, 2 cm in diameter, is placed in the xy-plane with its center at P. The magnetic flux through the coil is closest to:

a)4.9 x 10-9 Wb
b)9.9 x 10-9 Wb
c)4.0 x 10-9 Wb
d)1.5 x 10-9 Wb
e)2.0 x 10-9 Wb

2. Homework Equations

[tex]B=\frac{\mu_{0}Ia^2}{2(x^2+a^2)^{3/2}}[/tex] (on the axis of a circular loop)


[tex]B=\frac{\mu_{0}NI}{2a}[/tex] (at the center of N circular loops)


[tex]\Phi=\int[/tex]B*dA (magnetic flux)



3. The Attempt at a Solution

So I tried finding the magnetic flux of the loop first in the image given. First I needed the field of the loop:

Using the first formula, I used I = 5 A, x = .25 m, a = .01 m. My final answer resulted in : 2.01E-8 T

Then, I used the formula for magnetic flux, using 2.01E-8 T for B, and the area of this circle.

Area of circle: [tex]2\pi*r^2[/tex], where r will be .01
[tex]2\pi*.01^2[/tex] = 6.28E-4

so

[tex]\Phi[/tex] = 2.01E-8 * 6.28 E-4 = 1.26 E -11

***

I figure that somehow, I needed the magnetic flux of the loop to figure out what the flux of the coil would be. Am I even approaching this correctly?
 

Answers and Replies

Shooting Star
Homework Helper
1,975
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3. The Attempt at a Solution

So I tried finding the magnetic flux of the loop first in the image given. First I needed the field of the loop:

Using the first formula, I used I = 5 A, x = .25 m, a = .01 m. My final answer resulted in : 2.01E-8 T
You should put x=radius of the loop, a=0. Do you know what x and a represent? Consult your notes or book.

Where is the field due to the straight wire? You have to include that too.

After finding the total B at the centre, think about finding the flux.
 

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