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pogonyaev
3 years ago
6

You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimen

sions of the coil go from a = 8 cm and b = 17 cm, to a* = 16 cm and b* = 22 cm in t = 0.04 seconds. If the coil has resistance that remains constant at 1.2 ohms. What would be the magnitude of the induced current in amperes?
Physics
1 answer:
taurus [48]3 years ago
7 0

Answer:

The  current is I  =  0.5425 \ A

Explanation:

From the question we are told that

   The  magnetic field is  B  =  1.2 \ T

   The first length is  a =  8 \ cm  =  0.08 \ m

    The  second length is  a^*  = 16 \ cm  =  0.16 \ m

    The first width is  b  =  17 \ cm  =  0.17 \ m

     The second  width is  b^*  =  22 \ cm  =  0.22 \ m

    The time interval  is  dt =  0.04 \ s

     The resistance is  R =  1.2 \ \Omega

Generally the first area is

     A =  a * b

=>    A =  0.08 *  0.17

=>     A =  0.0136  \  m^2

The second area is  

      A^*  =  a^*  * b^*

=>   A^*  =  0.16 * 0.22

=>     A^*  =  0.0352 \ m^2

Generally the induced emf is mathematically represented as

       \epsilon  = -  \frac{ B  * [A^*  -  A]}{dt}

This negative show that it is moving in the opposite direction of the motion producing it

=>   |\epsilon | =  \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}

=>    |\epsilon |  =  0.651 \ V

The induced current is

     I  =  \frac{|\epsilon|}{R}

=>   I  =  \frac{ 0.651}{1.2}

=>   I  =  0.5425 \ A

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