I need help with this physics question. Problem 20.33 4019 Constants Part A A lo
ID: 1576828 • Letter: I
Question
I need help with this physics question.
Problem 20.33 4019 Constants Part A A long horizontal wire carries 22.0 A of current due north. The Earth's field 25.0 cm due west of the wire points north but downward, 46°below the horizontal, and has magnitude of 5.0 × 10-5 T What is the magnitude of the net magnetic field 25.0 cm due west of the wire? Express your answer using two significant figures. B= 3.gx 10-5 T. Prevlous Answers Correct Part B What is the angle below the horizontal of the net magnetic field 25.0 cm due west of the wire? Express your answer using two significant figures. 25 below the horizontal Submit Previous Answers Request Answer Incorrect; One attempt remaining; Try AgainExplanation / Answer
A)
at 25 cm distance magnetic field due to the wire,
B = mue*I/(2*pi*r)
= 4*pi*10^-7*22/(2*pi*0.25)
= 1.76*10^-5 T
Bh = 5*10^-5 T
angle between B and Bh = 90 + 46
= 136 degrees
Bnet = sqrt(B^2 + Bh^2 + 2*B*Bh*cos(theta))
= sqrt(1.76^2 + 5^2 + 2*1.76*5*cos(136))
= 3.93*10^-5 T <----Answer
B) Bnety = B - Bh*sin(46)
= 1.76*10^-5 - 5*10^-5*sin(46)
= -1.84*10^-5 T
Bnetx = Bh*cos(46)
= 5*10^-5*cos(46)
= 3.47*10^-5 T
theta = tan^-1(Bnety/Bnetx)
= tan^-1(1.84/3.47)
= 28 degrees below the horizontal <----Answer