Constants Part A A long horizontal wire carries 25.0 A of current due north The

Question

Constants Part A A long horizontal wire carries 25.0 A of current due north The Earth's field 25.0 cm due west of the wire points north but downward, 47° below the horizontal, and has magnitude of 5.0x10 3T What is the magnitude of the net magnetic field 25.0 cm due west of the wire? Express your answer using two significant figures. Submit Request Answer Part B What is the angle below the horizontal of the net magnetic Express your answer using two significant figures. field 25.0 em due west of the wire? o below the horizontal

Explanation / Answer

a)
magnetic field due to the wire,

B1 = mue*I/(2*pi*r)

= 4*pi*10^-7*25/(2*pi*0.25)

= 2*10^-5 T (upward)

earth magnetic field, B2 = 5*10^-5 T (below 47 degrees)

Bnet = sqrt(B1^2 +B2^2 + 2*B1*B2*cos(theta))

= sqrt(2^2 + 5^2 + 2*2*5*cos(90 + 47))*10^-5

= 3.79*10^-5 T <<<<<<----Answer

b) Bx = B2*cos(47)

= 5*10^-5*cos(47)

= 3.41*10^-5 T

By = B1 - B2*sin(47)

= (2 - 5*sin(47))*10^-5

= -1.66*10^-5 T

theta = tan^-1(By/Bx)

= tan^-1(1.66/3.41)

= 26.0 degrees below horizontal <<<<<<<<----Answer

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