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Magnitism Question-help!

  • #1

Homework Statement



The magnitude of the magnetic force between two conductors 4.00m long, separated by 8.00cm is, 2.85X10^-5 N. The current in one conductor is twice the current in the other. Calculate both currents.

Given: FM = 2.85X10^-5 N
l = 4.00m
Mo = 4pie X10^-7
d = 0.08m
I2 = 2(I1)
I1 = ?

Homework Equations


F2/l = Mo(I1)(I2)/2pie(d)



The Attempt at a Solution


Couldnt really understand how I am supposed to find one of the currents when i dont know either of them, if i knew one i could find it easily.
 

Answers and Replies

  • #2
rock.freak667
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The magnitude of the magnetic force between two conductors 4.00m long, separated by 8.00cm is, 2.85X10^-5 N. The current in one conductor is twice the current in the other. Calculate both currents.
Call one current [itex]I_1[/itex] and the other current [itex]I_2[/itex]

Couldnt really understand how I am supposed to find one of the currents when i dont know either of them, if i knew one i could find it easily.
They directly told you how to relate one current to the other.
 
  • #3
Call one current [itex]I_1[/itex] and the other current [itex]I_2[/itex]



They directly told you how to relate one current to the other.
Um yes they told me but im not sure how im supposed to use that in the equation when i still dont know either of the actual currents. How am I supposed to set up the equation knowing one current is twice the other current?
 
  • #4
rock.freak667
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If

[tex]F=\frac{\mu_0 I_1 I_2 l}{2 \pi d}[/tex]

and that [itex]I_1=2I_2[/itex]

can you substitute that into the equation and then solve?
 
  • #5
Well all I can get from that is that it would be Mo2(I2)(I2)(l)/2pie(d) and the I2's would cancel out so ur left with F = Mo(2)(l)/2pie(d) but then how can i solve for I? I put in the values into that equation and did not end up with the correct answer
 
  • #6
rock.freak667
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if [itex]I_1=2I_2[/itex] and you put that into the formula, [itex]I_2 \times I_2 =(I_2)^2[/itex]
 
  • #7
Ok so then F=Mo(2)(I2)^2(l)/2pie(d) plug in 2.85X10^-5 for F, plug in 4pieX10^-7 for Mo, plug in 4 for l and plug in 0.08 for d. I calculated and ended up with 1.19 X10^-7, the answer is supposed to be 1.18 and 2.36
 
  • #8
rock.freak667
Homework Helper
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Ok so then F=Mo(2)(I2)^2(l)/2pie(d) plug in 2.85X10^-5 for F, plug in 4pieX10^-7 for Mo, plug in 4 for l and plug in 0.08 for d. I calculated and ended up with 1.19 X10^-7, the answer is supposed to be 1.18 and 2.36
Then you plugged in the values wrongly or calculated wrong. I got close to those answers.
Make [itex](I_2)^2[/itex] the subject and then plug in the values.
 
  • #9
Ok I tried that so its (I2)^2 on one side so it would be = to F(2pie x d)/Mo(2)(l) which still equalled 1.19X10^-7 so i dont know how or what way u used to get it
 
  • #10
rock.freak667
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[tex]\frac{2\pi Fd}{2 \mu l}= \frac{2.85 \times 10^{-5} \times 2\pi \times 0.08}{2\times 4\pi \times 10^{-7} \times 4}[/tex]

Did you find that?
 
Last edited:
  • #11
That equals 2.85
 
  • #12
rock.freak667
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