A coil formed by wrapping 65 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 36.0° with the direction of the field. When the magnetic field is increased uniformly from 200 µT to 600 µT in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire?(HINT: find the Area of the square coil first, and use that to find the length of each side (s). The total length of the coil is the number of turns times 4 times the length of the side: L=N(4s)

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Vespertilio Vespertilio · Answer 1 · 2019-12-15T13:58:52+01:00

Answer:

377 m

Explanation:

number of turns, N = 65

θ = 36°

B1 = 200 micro Tesla

B2 = 600 micro tesla

t = 0.4 s

induced emf, e = 80 mV

Let a be the side of the square coil.

[tex]e=\frac{d\phi }{dt}=NA\frac{dB}{dt}\times Sinθ[/tex]

[tex]0.080=\frac{65\times a^{2}\times Sin36\times\left ( 600 - 200 \right )\times 10^{-6}}{0.4}[/tex]

[tex]0.080=0.038a^{2}[/tex]

a = 1.45 m

Total length of the wire, L = N x 4a = 65 x 4 x 1.45 = 377 m

Thus, the length of the wire is 377 m.