# Homework Help: Magnitism Question-help!

1. May 21, 2008

1. The problem statement, all variables and given/known data

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 = ?

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

3. 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.

2. May 21, 2008

### rock.freak667

Call one current $I_1$ and the other current $I_2$

They directly told you how to relate one current to the other.

3. May 21, 2008

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. May 21, 2008

### rock.freak667

If

$$F=\frac{\mu_0 I_1 I_2 l}{2 \pi d}$$

and that $I_1=2I_2$

can you substitute that into the equation and then solve?

5. May 21, 2008

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. May 21, 2008

### rock.freak667

if $I_1=2I_2$ and you put that into the formula, $I_2 \times I_2 =(I_2)^2$

7. May 21, 2008

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. May 21, 2008

### rock.freak667

Then you plugged in the values wrongly or calculated wrong. I got close to those answers.
Make $(I_2)^2$ the subject and then plug in the values.

9. May 21, 2008

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. May 21, 2008

### rock.freak667

$$\frac{2\pi Fd}{2 \mu l}= \frac{2.85 \times 10^{-5} \times 2\pi \times 0.08}{2\times 4\pi \times 10^{-7} \times 4}$$

Did you find that?

Last edited: May 21, 2008
11. May 21, 2008