Hw 05 13 of 41 Problem 20.35 Part A Constants A long horizontal wire carries 19.

Question

Hw 05 13 of 41 Problem 20.35 Part A Constants A long horizontal wire carries 19.0 A of current due north what is the absolute value of te net magnetic field260cm due west of the wro ette Earth's neid there points north butdownward, 43 below the horizontal, and has magntude 5.0 × 10 'T? Express your answer using two significant figures 4.5-10 Submit Previous Answers Request Answer Incorrect: Try Again PartB what is the angle of the net magnetic field 26.0 cm due west of te wie if the Earth's feld thro ports north but downward,43° below the horizontal, and has magnitude 5.0 x10 T? Express your answer using two significant figures below the horizontal

Explanation / Answer

A)

Magnetic field of a long straight wire: B = o*I/2R

o = 4×10^-7 Tm/A

I = 19 A

R = 26cm = 0.26 m

Be = 5×10^-5 T at -43°

The magnetic field of the wire due west is directed straight up, using the

right hand rule.

Bw = ((4×10^-7)*19) / (2*0.26) = 1.46×10^-5 T

The Earth's magnetic field points below the horizontal at this location.

Components of the Earth's field:

Bex = (5×10^-5)cos(-43°) = (5×10^-5)cos43°

Bey = (5×10^-5)sin(-43°) = -(5×10^-5)sin43°

Total field in the x-direction: Bx = Bex = (5×10^-5)cos43°

Total field in the y-direction: By = Bw+Bey = (1.46×10^-5)-(5×10^-5)sin43°

Net magnetic field: B = (Bx^2+By^2)

B = [ ((5×10^-5)cos43°)^2 + ((1.46×10^-5)-(5×10^-5)sin43°)^2 ]

B = 4.1×10^-5 T

B)

tan = By/Bx

= arctan (By/Bx)

= 28° below the horizontal

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