Magnitude and Direction of Magnetic Field

AI Thread Summary
The discussion focuses on calculating the magnetic field at point A due to two crossed wires carrying a current of 3.4 A each. The initial calculation yielded a magnetic field of 5.7 x 10^-8 T, but it was pointed out that the distances should be expressed in standard units. After correcting the distances to meters, the final magnetic field value was confirmed as 5.7 x 10^-6 T directed out of the page. The conversation highlights the importance of unit conversion in physics calculations. The final answer is now validated and accurate.
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Homework Statement


Two long straight wires each carry 3.4 A of current and are crossed, as shown in the figure. If Dh=12.0cm and Dv=6.0cm, what is the magnitude and direction of the magnetic field at A? Figure: http://i41.tinypic.com/4louv6.jpg"

Homework Equations


B=(Mu)(I)/(2pie)(r)

The Attempt at a Solution


B = [(4pie x 10^-7 Tm/A)(3.4A)]/(2pie) x (1/6.0cm - 1/12.0cm)
B = 5.7 x 10^-8 T out of page

Would someone be able to please check if that is the correct answer? Thank you!
 
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hardwork said:
B = [(4pie x 10^-7 Tm/A)(3.4A)]/(2pie) x (1/6.0cm - 1/12.0cm)
B = 5.7 x 10^-8 T out of page
Almost. Express those distances in standard units.
 
Oh, right. Thank you!

B = 5.7 x 10^-6 T out of page
 
hardwork said:
B = 5.7 x 10^-6 T out of page
Good!
 

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