Question

One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 5.87 × 10-3 m2. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 14.9 Ω. If an externally applied uniform longitudinal magnetic field in the core changes from 1.93 T in one direction to 1.93 T in the opposite direction, how much charge flows through a point in the circuit during the change?

Answer 1

Answer:

The amount of charge that flow through a point is $\Delta q = 0.126 \ C$

Explanation:

From the question we are told that

    The number of turns is  $N = 100$

    The area is  $A = 5.87 *10^{-3} m^2$

       The total resistance is $R = 14.9 \ \Omega$

       The magnetic field in first direction  is  $B_1 = 1.93 \ T$

        The magnetic field in second   direction  is $B_2 = -1.93 \ T$

The change in magnetic field is evaluated as

     $\Delta B = B_1 - B_2$

substituting values

      $\Delta B = 1.93 - [- 1.93]$

       $\Delta B =3.2 \ T$

The induced emf due to the change is mathematically evaluated as

        $e = NA \frac{\Delta B }{\Delta t }$

This can also be mathematically represented as

     $e = IR$

     Where I can be mathematically represented as

     $I = \frac{\Delta q }{\Delta t}$

So  

     $e = \frac{\Delta q}{\Delta t } R$

Now  

      $\frac{\Delta q}{\Delta t } R = NA \frac{\Delta B }{\Delta t }$

=>   $\Delta q = \frac{N A (\Delta B)}{R}$

substituting values

     $\Delta q = \frac{100 * 5.87 *10^{-3} (3.2}{14.9}$

     $\Delta q = 0.126 \ C$