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Magnetic Field in a Rectangular Conducting Loop

  • Thread starter daimoku
  • Start date
20
0
[SOLVED] Magnetic Field in a Rectangular Conducting Loop

1. Homework Statement
Figure 31-64a shows a rectangular conducting loop of resistance R = 0.010 , height H = 1.5 cm, and length D = 2.5 cm being pulled at constant speed v = 55 cm/s through two regions of uniform magnetic field. Figure 31-64b gives the current i induced in the loop as a function of the position x of the right side of the loop. For example, a current of 3.0 µA is induced clockwise as the loop enters region 1. What are the magnitudes and directions of the magnetic field in region 1 and region 2?

http://personalpages.tds.net/~locowise/test/W0736-N.jpg


2. Homework Equations
[tex] EMF=BLV [/tex]

[tex] i=EMF/R [/tex]



3. The Attempt at a Solution
Okay, I found the magnetic field in region 1 like so:
3*10^-6A = EMF / 0.010 ohms
EMF=3*10^-8 V

3*10^-8 V = B * 0.015m * 0.55 m/s
B = 3.64 microT for region 1

However, for region 2 I must be making a mistake somewhere. Could someone point out my mistake please? Here's what I tried:

-2*10^-6A = EMF/ 0.010 ohms
EMF = 2*10^-8 V
EMF=BLV
2*10^-8V = B * 0.55 m/s * 0.015m
B=2.42 microT

I have a feeling the mistake is in the length but I don't entirely understand how to interpret the problem statement. Thanks for your help!
 
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Answers and Replies

20
0
I tried taking the difference between magnetic fields 1 and 2 but it doesn't seem to be working. Can anyone offer some help?
 
Re: [SOLVED] Magnetic Field in a Rectangular Conducting Loop

I realize this post is over two years old, but for anybody who cannot figure out this problem and stumbles across this page, here is the answer for region 2.

EMF(2) = i(1)R + i(2)R = R[ i(1) + i(2) ] = (0.010 ohm)[(3*10^-6A) + (-2*10^-6A)]
EMF(2) = 1*10^-8V

Then use the same equation that daimoku was using to find B(2) for region 2.

B(2) = EMF(2) / (v * H) = (1*10^-8) / (0.55m/s * 0.015m) = 1.21 microT.

Both B(1) and B(2) are out of the page.
 

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