Question

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 dimensions 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?

Answer 1

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

     $I = \frac{|\epsilon|}{R}$

=>   $I = \frac{ 0.651}{1.2}$

=>   $I = 0.5425 \ A$