A long horizontal wire carries 28.0 A of current due north. Part A What is the n

Question

A long horizontal wire carries 28.0 A of current due north.

Part A What is the net magnetic field 21.0 cm due west of the wire if the Earth's field there points downward, 44° below the horizontal, and has magnitude 5.0 x 105T? Express your answer using two significant figures. B 8.88 10 Submit Previous Answers Request Answer X Incorrect: Try Again; 4 attempts remaining Part B Express your answer using two significant figures. 0-66.17 o below the horizontal Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

componet of earth magnetic field

BEz = -5*10^-5*sin44 = 3.47*10^-5 T ( into of page)

BEy = 5*10^-5*cos44 = 3.6*10^-5 T ( north)

magnetic field due to current carrying wire B1z = uo*I/(2*pi*r) ( out of page)

B1z = 4*pi*10^-7*28/(2*pi*0.21) = 2.67*10^-5 T

Bz = BEz + B1z = -0.8*10^-5 T

By = 3.6*10^-5 T

magnitude of net magnetic field B = sqrt(By^2+Bz^2) = 3.7 *10^-5 T

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part B

direction = tan^-1(By/Bz) = 77 degrees

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