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jeyben [28]
3 years ago
13

A 123-turn circular coil of radius 2.41 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi

l. During 0.151 s the magnetic field strength increases from 50.9 mT to 96.3 mT. Find the magnitude of the average EMF, in millivolts, that is induced in the coil during this time interval.
Physics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

67.44 V

Explanation:

Number of turns N =123

Radius = 2.41 cm =0.0241 m

The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T

Time interval dt = 0.151 sec

We know that the induced emf e=-NA\frac{dB}{dt}

Area A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2

Putting all these values in emf equation e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V here negative sign indicates that it opposes the cause due to which it is produced

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When is the magnitude of the acceleration of a mass on a spring at its maximum value?
Andreas93 [3]

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In fact:

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- When the displacement of the spring is maximum, the velocity is zero, due to conservation of energy. In fact, as the displacement is maximum, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is maximum, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be zero, and so the velocity (v) is also zero. On the cotnrary, since acceleration (a) is directly proportional to the restoring force of the spring, given by

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Based on this, the correct answer is

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5 0
2 years ago
Read 2 more answers
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