Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70

Answer 1

Answer:

The  induced current is $I = 6.25*10^{-4} \ A$

Explanation:

From the question we are told that  

    The number of turns is  $N = 1$

     The  cross-sectional area is  $A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2$

    The  initial magnetic field is  $B_i = 0.500 \ T$

     The  magnetic field at time =  1.02 s  is  $B_t = 2.60 \ T$

     The  resistance is  $R = 2.70\ \Omega$

The  induced emf is mathematically represented as

       $\epsilon = - N * \frac{ d\phi }{dt}$

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  $d\phi$ is the change in magnetic flux which is mathematically represented as

        $d \phi = dB * A$

Where  dB  is the change in magnetic field which is mathematically represented as

        $dB = B_t - B_i$

substituting values

        $dB = 2.60 - 0.500$

        $dB = 2.1 \ T$

Thus  

      $d \phi = 2.1 * 8.20 *10^{-4}$

     $d \phi = 1.722*10^{-3} \ weber$

So  

     $|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}$

     $|\epsilon| = 1.69 *10^{-3} \ V$

The  induced current i mathematically represented as

      $I = \frac{\epsilon}{ R }$

  substituting values

       $I = \frac{1.69*10^{-3}}{ 2.70 }$

       $I = 6.25*10^{-4} \ A$