Magnetic field of two current-carrying wires crossing over

[ness87]

Two long current-carrying wires cross at an angle of 21° ("theta" is half of this) as shown in the figure above. The magnitude of the current is the same in each wire, I=400 A.

A wood mouse is running along the dashed line midway between the wires towards the point where the wires cross. The mouse turns back at point P, some distance x from the wire crossing point, because the magnetic field strength reaches an unbearable 6.8 mT. Determine the distance x (in cm).


Okay, so I assigned the dotted line perpendicular to one of the wires (in the triangle of interest in the diagram) as 'r' and...

I did:

B=(μ0/2pi) x I/r
so r =μ0/2pi x I/B
Therefore
r=(2x10^(-7) x 400)/6.8
Therefore r = 1.176x10^(-5)

Now using sine
sin(theta) = r/x
sin(10.5)=(1.176x10^(-5))/x
therefore x = (1.176x10^(-5))/(sin(10.5))
=6.45 x 10^(-5)
=0.0000645meters
=0.00645cm
or 6.45x10^(-3)cm

WHICH IS WRONG!
The weird thing is if i multiply that by two and move the decimal i have 12.9cm which is close to the correct answer of 12.91cm. Maybe I have stuffed up the formula or the units somewhere...

Also I'm aware the question is in millitesla so maybe i should have used 0.0068T in my first equation. However when I compute this I still get the wrong answer.

If you can help that would be great, I thought I was doing this the right way but maybe there is another way.

 

[Sleepy_time]

Hi ness87. The distance from both wires at P are equal and the fields due to both are the sum of their individual fields. So the field would be $B=2\frac{\mu_0I}{2{\pi}r}$. Using $B=6.8\times10^{-3}$ T also, gives you a factor of 2000 off from the answer

 

[ness87]

Ahhh! Fantastic, yes that makes sense to do twice the field strength because both fields contribute. Great!

Thank you very much sleepy time, much appreciated