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alex41 [277]
3 years ago
10

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 6.40×10−5 T . Part A
What is the total magnitude of the magnetic flux ( ?initial) through the coil before it is rotated?

Express your answer numerically, in webers, to at least three significant figures.

Part B

What is the magnitude of the total magnetic flux ?final through the coil after it is rotated?

Express your answer numerically, in webers, to at least three significant figures.

Part C

What is the magnitude of the average emf induced in the coil?

Express your answer numerically (in volts) to at least three significant figures.
Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

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