(50%) Problem 2: Two 0.35 m long parallel wires are separated by a distance d =

Question

(50%) Problem 2: Two 0.35 m long parallel wires are separated by a distance d = 0.15 m and carry a current I= 1.5 A., as shown in the figure, the currents in the wires are flowing in the same direction What is the magnitude of the magnetic potential (in W/Am) of this system at a position P half-way in-between the two wires along the perpendicular biscctor of the wires? Grade Summary Deductions Potential 0% 100% cos) cotan0asin0 acos0 atan)acotanOsinh) coshO tanh) cotanhO Degrees Radians tan() | | Submissions in Attempts remaining: 20 (0% per attempt) detailed view 12 3 0 END DEL CLEAR Submit int I give up! Hints: 0 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback.

Explanation / Answer

given

l = 0.35 m
d = 0.15 m
I = 1.5 A

now, magnetic field at a point where ditance form the lower wire is x

Bx = 2kI*l/2(d - x)*sqrt(l^2/4 + (d - x)^2) + 2kI*l/2x*sqrt(l^2/4 + x^2)
Bx = 2kI*l/(d - x)*sqrt(l^2 + 4(d - x)^2) + 2kI*l/x*sqrt(l^2 + 4x^2)

magnetic al halfways is then V
V = integrate Bx*dx from x = 0 to x = d/2
V = 2kI[ln(l*sqrt(4(d - d/2)^2 + l^2) + l^2) - ln(d - d/2)] - 2kI[ln(l*sqrt(4(d)^2 + l^2) + l^2) - ln(d)] + 2kI*(ln(d/2) - ln(l(sqrt(l^2 + 4d^2/4) + l))) - 2kI*(ln(0) - ln(l(sqrt(l^2) + l)))

V = 2kI[ln(l*sqrt(d^2 + l^2) + l^2) - ln(d/2)] - 2kI[ln(l*sqrt(4d^2 + l^2) + l^2) - ln(d)] + 2kI*(ln(d/2) - ln(l(sqrt(l^2 + 4d^2/4) + l)))
V = 2kI(1.226814 - 0.63778 + 0.379516) = 0.209565 uJ/Am

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