Question

A flat loop of wire consisting of a single turn of cross-sectional area 7.10 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.10 T in 1.07 s. What is the resulting induced current if the loop has a resistance of 1.60?

Answer 1

Answer:

The induced current is  $I = 0.00066 \ A$

Explanation:

From the question we are told that

   The area is  $A = 7.10 \ cm^2 = 7.10 *10^{-4} \ m^2$

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

    The magnetic field after t =1.07 s is  $B_f = 2.10 \ T$

     The resistance of the loop is $R = 1.60 \ \Omega$

Generally the electromagnetic field induced is mathematically represented as

   $\epsilon = NA * \frac{B_f - B_i}{t}$

Where N is the number of turns which is 1 in the case of this question since there is only one loop

 So

       $\epsilon = 1 * 7.10*10^{-4}* \frac{2.10 - 0.500}{1.07 }$

=>   $\epsilon = 0.00106 \ V$

Generally the value of the current is mathematically represented as

        $I = \frac{\epsilon}{R}$

       $I = \frac{0.00106}{1.60}$

       $I = 0.00066 \ A$