(1896) Problem 1: Consider two current-carrying wires, separated by a distance d
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Question
(1896) Problem 1: Consider two current-carrying wires, separated by a distance d= 5.5 cm, as shown in the figure. The left wire is directed out of the page with current 11, and the right wire is directed into the page with current I2. The point P is a distance d from both wires, so the wires and the point form an equilateral triangle Assignment Status Click here for detailed view Otheexpertta.com Problem Status Partial 50% Part (a) lfboth wires are carrying a current of 9.5 A. what is the magnitude of the magnetic field in tesla at point p? 50% Part (b) If the current from the first wire is 9.5 A and the current from the second is 14.5 A, what is the magnitude of the magnetic Completed field, in tesla, at point P? Grade Summary Deductions Potential Partial 0% 100% 6 Completed cos0 cotanasin acosO atan0acotansinhO cosh tanh0cotanh( Submissions Attempts remaining: 8 (3 per attempt) detailed view Degrees Submit Hint I give up! Hines: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. Submission History Hints Feedback Totals Totals 0% 096 6 06Explanation / Answer
a) magnitude of magnetic field due to each wire at the location of P,
B1 = B2 = mue*I/(2*pi*d)
= 4*pi*10^-7*9.5/(2*pi*0.055)
= 3.45*10^-5 T
the angle between B1 and B2 is 120 degrees
so,
Bnet = sqrt(B1^2 + B2^2 + 2*B1*B2*cos(120))
= B1 (since |B1| = B2| )
= 3.45*10^-5 T
b) B1 = mue*I/(2*pi*d)
= 4*pi*10^-7*9.5/(2*pi*0.055)
= 3.45*10^-5 T
B2 = mue*I2/(2*pi*d)
= 4*pi*10^-7*14.5/(2*pi*0.055)
= 5.27*10^-5 T
Bnet = sqrt(B1^2 + B2^2 + 2*B1*B2*cos(120))
= sqrt(3.45^2 + 5.27^2 + 2*3.45*5.27*cos(120))*10^-5
= 4.64*10^-5 T
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