Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 G to 1.60 M in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20 Ω?

Answer 1

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of $1.20 \ \Omega$

Answer:

The current is   $I = 0.0007 41 \ A$

Explanation:

From the question we are told that

   The  area is  $A = 8.00 \ cm^2 = 8.0 *10^{-4} \ m^2$

   The initial magnetic field at $t_o = 0 \ seconds$  is $B_i = 0.500 \ T$

   The magnetic field at $t_1 = 0.99 \ seconds$ is  $B_f = 1.60 \ T$

     The resistance is  $R = 1.20 \ \Omega$

Generally the induced emf is mathematically represented as

      $\epsilon = A * \frac{B_f - B_i }{ t_f - t_o }$

=>   $\epsilon = 8.0 *10^{-4} * \frac{1.60 - 0.500 }{ 0.99- 0 }$

=>   $\epsilon = 0.000889 \ V$

Generally the current induced is mathematically represented as

     $I = \frac{\epsilon}{R }$

=>  $I = \frac{0.000889}{ 1.20 }$  

=>  $I = 0.0007 41 \ A$