Magnitism Question-help

[Bradsteeves]

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 don't know either of them, if i knew one i could find it easily.

 

[rock.freak667]

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 $I_1$ and the other current $I_2$

Couldnt really understand how I am supposed to find one of the currents when i don't 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.

 

[Bradsteeves]

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



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

Um yes they told me but I am not sure how I am supposed to use that in the equation when i still don't know either of the actual currents. How am I supposed to set up the equation knowing one current is twice the other current?

 

[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?

 

[Bradsteeves]

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

 

[rock.freak667]

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

 

[Bradsteeves]

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

 

[rock.freak667]

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 $(I_2)^2$ the subject and then plug in the values.

 

[Bradsteeves]

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 don't know how or what way u used to get it

 

[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?

 

[Bradsteeves]

That equals 2.85

 

[rock.freak667]

That equals 2.85

No I get 1.425