A long horizontal wire carries 16.0 A of current due north. A. What is the absol

Question

A long horizontal wire carries 16.0 A of current due north.

A. What is the absolute value of the net magnetic field 21.0 cm due west of the wire if the Earth's field there points north but downward, 47degrees below the horizontal, and has magnitude 5.0 times 10^-5 T?

B. What is the angle of the net magnetic field 21.0 cm due west of the wire if the Earth's field there points north but downward, 47degrees below the horizontal, and has magnitude 5.0 times 10^- 5?

I'm not sure where to start with this....any help would be great, thank you!

Explanation / Answer

Magnetic field due to the wire B = _0 I / 2r Here I = 16.0 A r = 0.21 m B_wire (pointing upward) = (4*10^-7 T.A/m)(16.0A)/ 2(0.21m) = 1.52*10^-5 T magnetic filed due to earth B_Ex = (5*10^-5 T )cos(360 -47)^0   = (5*10^-5 T )cos313 B_Ex = 3.405*10^-5 T B_Ey = (5*10^-5 T )sin(360 -47)^0   =   (5*10^-5 T )sin313 = -3.656 *10^-5 T Net field components B x =  3.405*10^-5 T B_y = B_wire   + B_Ey =  - 2.136 *10^-5 T Net field = B = sqrt( (B_x)^2 + (B_y)^2) B = 4.019 *10^-5 T direction = tan^-1 (B_y / B_x)   = 32.1 ^0 since y component is negative and x component is positive The direction of net magnetic field is 32.1^0 clock wise from the positive X axis B_y = B_wire   + B_Ey =  - 2.136 *10^-5 T Net field = B = sqrt( (B_x)^2 + (B_y)^2) B = 4.019 *10^-5 T direction = tan^-1 (B_y / B_x)   = 32.1 ^0 since y component is negative and x component is positive The direction of net magnetic field is 32.1^0 clock wise from the positive X axis

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