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flyingpig
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Homework Statement
Two long parallel conductors separated by 10.0 cm carry currents in the same direction. The first wire carries a current I1 = 5.00 A, and the second carries I2 = 8.00 A.
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(a) What is the magnitude of the magnetic field created by I1 and at the location of I2
(b) What is the force per unit length exerted by I on I2?
The Attempt at a Solution
For (a) it is
[tex]\left |\vec{B_1} | \right= \frac{\mu_{0} I_{1}}{2\pi d}[/tex]
Then it is just a matter of plugging in the numbers and it should come out as 10-5T
But my question is, why is it that we use r = 10.00cm? Shouldn't it be d + x?
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Now for (b), I am just confused with Newton's third Law. I just don't understand I am wrong.
Look at picture
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By the right hand rule - cross product - the force F1 should be to the right, that is the Force exerted by wire 1 (which I forgot to label as the one on the right) on wire 2.
If I were to do the Math I get
[tex]\left | \vec{F_{12}} \right | = I_{1} l \left| \vec{B_1} \right| [/tex]
[tex]\left | \vec{F_{12}} \right | = I_{1} \frac{\mu_{0} I_{1}}{2\pi d}[/tex]
[tex]\frac{\left | \vec{F_{12}} \right |}{l} = \frac{\mu_{0} I_{1}^2}{2\pi d}[/tex]
Which is wrong, but it should be [tex]\frac{\left | \vec{F_{12}} \right |}{l} = \frac{\mu_{0} I_{1} I_{2}}{2\pi d}[/tex] by Newton's third law
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