Posted by **Robert** on Sunday, August 12, 2012 at 11:12pm.

Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire?

So I figured out that it's a triangle that forms with 25cm as the hypotenuse (distance between wire 2 and B)

From here I'll use B = mu_0 I / 4pi r

For Wire 1 I get 20 mu_T

For wire 2 I get 16 mu_T

so together I get 36 mu_T

Did I do this right?

Thank you

- Physics - Verification -
**Elena**, Monday, August 13, 2012 at 3:03am
B1=μₒ•I1/2•π•r1 =4π•10^-7•30/2•π•0.15 =4•10^-5 T,

B2=μₒ•I2/2•π•r2 =4π•10^-7•40/2•π•0.25 =3.2•10^-5 T.

Generally there are two points which are separated by given distances from I1 (point M) and I2 (point N), where we have to find magnetic field B. They are above and below MN line.

Let us examine the point P which is below the line connecting I1 and I2.

Let the left current I1 be directed into the page,

and the right current I2 be directed out of the page.

B1 is normal to MP line and is directed due to west,

B2 is normal to NP line and is directed south east.

Net B is directed southwest.

The magnutude of B may be found using cosine law.

The angle α is equal to the angle MPN which cosine is

cosα=15/25=0.6.

B=sqrt(B1²+B2²-2B1•B2•cosα) =

=10^-5•sqrt[ (4²+3.2²-2•4•3.2•0.6)]= 3.3•10^-5 T.

If I1 is out of the page and I2 is into the page,

net B is directed northeast and is of the same magnitude.

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