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A long thin wire carrying a current I1=10A (out of page) is located at (0,0) and...

A long thin wire carrying a current I1=10A (out of page) is located at (0,0) and another wire carrying a current I2=8A (into the page) is located at (x=9cm,y=0). Find the magnitude and direction the magnetic field due to these two currents at (x=0,y=12 cm). Your solution must include a neat diagram.

Solutions

Expert Solution

B1 = magnetic field by current I1 = (/4) (2I1/AP) = (10-7) (2 (10/(0.12))) = 1.67 x 10-5 T

B2 = magnetic field by current I2 = (/4) (2I2/BP) = (10-7) (2 (8/(0.15))) = 1.1 x 10-5 T

Net magnetic field along X-direction is given as

Bx = B2 Sin - B1 = (1.1 x 10-5) Sin53.12 - (1.67 x 10-5) = - 7.9 x 10-6 T

Net magnetic field along Y-direction is given as

By = B2 Cos  = (1.1 x 10-5) Cos53.12 = 6.6 x 10-6 T

Net magnetic field is given as

B = sqrt(Bx2 + By2) = sqrt((- 7.9 x 10-6)2 + (6.6 x 10-6)2) = 1.03 x 10-5 T

direction is given as

= 180 - tan-1(By/Bx) = 180 - tan-1(6.6/7.9) = 140 deg counterclockwise from positive X-axis

genius_generous answered 2 years ago
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