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LenKa [72]
2 years ago
10

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35m x 0.55m. The magnetic field has a

magnitude of 2.1T and is inclined at an angle of 65 degrees with respect to the normal to the plane of the loops. (a) If the magnetic field decreases to zero in a time of 0.45s, what is the magnitude of the average emf induced in the loop.(b) If the magnetic field remains constant at its initail calue of 2.1T, what is the magnitude of the rate ΔA/Δt at which the area should change so that the average emf has the same magnitude as in part (a)?

Physics
1 answer:
Cloud [144]2 years ago
4 0

Answer:

E = 0.38V

Explanation:

See attachment below.

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Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

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According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

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