Question

Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.

Answer 1

Answer:

a

  $\phi = 1.78 *10^{-7} \ Weber$

b

 $L = 1.183 *10^{-7} \ H$

Explanation:

From the question we are told that

   The radius is  $r = 6 \ cm = \frac{6}{100} = 0.06 \ m$

   The current it carries is  $I = 1.50 \ A$

     

The  magnetic flux of the coil is mathematically represented as

       $\phi = B * A$

Where  B is the  magnetic field which is mathematically represented as

         $B = \frac{\mu_o * I}{2 * r}$

Where  $\mu_o$ is the magnetic field with a constant value  $\mu_o = 4\pi * 10^{-7} N/A^2$

substituting  value

          $B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}$

          $B = 1.571 *10^{-5} \ T$

The area A is mathematically evaluated as

       $A = \pi r ^2$

substituting values

       $A = 3.142 * (0.06)^2$

       $A = 0.0113 m^2$

the magnetic flux is mathematically evaluated as    

        $\phi = 1.571 *10^{-5} * 0.0113$

         $\phi = 1.78 *10^{-7} \ Weber$

The self-inductance is evaluated as

       $L = \frac{\phi }{I}$

substituting values

        $L = \frac{1.78 *10^{-7} }{1.50 }$

         $L = 1.183 *10^{-7} \ H$