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maria [59]
3 years ago
12

A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a o

f the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.
Physics
1 answer:
Kruka [31]3 years ago
5 0

Answer:

0.2677\ \text{V/m}

Explanation:

A = Area of loop = 0.129\times0.402

B = Magnetic field = 0.888\ \text{T}

t = Time taken = 0.172\ \text{s}

Electric field is given by

E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}

The emf induced is 0.2677\ \text{V/m}.

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Read 2 more answers
a spring is compressed 15 centimeters when a 8.5 kilogram weight is set upon it. how much power would be needed to stretch out t
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Answer:

159.38 Watts

Explanation:

Initially;

  • Mass on the spring is 8.5 kg
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But;

F = kx

where F is the force of compression, k is the spring constant and x is the compression distance.

Thus;

k = F/x

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We are required to determine the power needed to stretch the same spring for 1.5 m in 4 secs.

Power = Work done ÷ time

Work done is given by 0.5kx²

Therefore;

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          = 159.38 Watts

Thus, the power needed is 159.38 watts

5 0
4 years ago
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