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Suppose a 50-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.250m2. It is stretched to have no area in 0.100 s. What is the direction and magnitude of the induced emf if the uniform magnetic field has a strength of 1.50 T?

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Answer:

187.5 V and the direction is counter clockwise

Explanation:

N = 50 turns, A1 = 0.25 m^2, A2 = 0, dt = 0.1 s

dA = A2 - A1 = 0.25 m^2, B = 1.5 T

induces emf, e = rate of change of magnetic flux linked with the coil

e = N x d (B A) / dt

e = N x B x dA/dt

e = 50 x 1.5 x 0.25 / 0.1 = 187.5 V

According to the Lenz law the direction of induced emf is such that it always opposes the responsible reason due to which it is produced.

So, the direction of induced emf is counter clock wise.

onyebuchinnaji

The magnitude of the induced emf in the uniform magnetic field is 187.5 V.

Induced emf in the magnetic field

The induced emf in the magnetic field is given as

emf = N(dφ)/dt

where;

  • N is the number of turns
  • φ is magnetic flux
  • t is time

emf = N(BA)/t

where;

  • B is magnetic field strength
  • A is the area of the coil

emf = (50 x 1.5 x 0.25)/(0.1)

emf = 187.5 V

Thus, the magnitude of the induced emf in the uniform magnetic field is 187.5 V.

Learn more about induced emf here: https://brainly.com/question/13744192

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