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Andrei [34K]
3 years ago
8

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

m 0.50 T to 1.50 T in 0.60 s. You may want to review (Pages 825 - 829) . For help with math skills, you may want to review: Rearrangement of Algebraic Expressions For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Electromagnetic induction. Part APart complete What is the induced emf in the coil?
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

The induced emf is  |\epsilon|=0.0261 V

Explanation:

From the question we are told that

     The number of turn is  N = 100

      The diameter of the coil is  d = 2.0 cm = 2.0 *10^{-2} m

      The uniform magnetic at initial is B_i = 0.50 T

      The uniform magnetic at initial is B_f = 1.50 T

      The time taken is t = 0.60s

      The angle the magnetic field makes with vertical is  \theta = 60^o

Generally induced emf is mathematically represented as

     \epsilon = -N \frac{d \o}{dt}

where d \o is the change  magnetic flux

Magnetic flux is mathematically represented as

       \O = \= B \cdot \= A

          = BA cos \theta

Substituting this above  

       \epsilon = -N \frac{d (BA cos \theta)}{dt}

      \epsilon = -N A \frac{d (B cos \theta)}{dt}

  Where B is the magnetic field and A is the area which is mathematically evaluated as

        A = \frac{\pi  d^2}{4}

Substituting values

        A = \frac{\pi (2.0 *10^{-2})^2}{4}

           A= 3.142*10^{-4}m^2

From the equation of emf

          \epsilon = -N A \frac{d (B cos \theta)}{dt}

dB = B_2 -B_1

       So

             |\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}

substituting values

            |\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}

                |\epsilon|=0.0261 V

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