Question

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .Part A. Calcualte the coil's self-inductance.Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf froma to b or from b to a?

Answer 1

Complete Question

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .

Part A. Calculate  the coil's self-inductance.

Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.

Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?

Answer:

Part A  

       $L = 0.000863 \ H$

Part B  

       $\epsilon = 0.863 \ V$

Part C

    From terminal a to terminal b

Explanation:

From the question we are told that

      The  number of turns is  $N = 590 \ turns$

      The cross-sectional area is  $A = 6.20 cm^2 = 6.20 *10^{-4} \ m$

      The  radius is $r = 5.0 \ cm = 0.05 \ m$

       

Generally the coils self -inductance is mathematically represented as

              $L = \frac{ \mu_o N^2 A }{2 \pi * r }$

Where $\mu_o$ is the permeability of  free space with value $\mu_o = 4\pi * 10^{-7} N/A^2$

substituting values

             $L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }$

             $L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }$

             $L = 0.000863 \ H$

Considering the Part B

      Initial current is $I_1 = 5.00 \ A$

      Current at time t is $I_t = 3.0 \ A$

       The  time taken is  $\Delta t = 3.00 ms = 0.003 \ s$

The self-induced emf is mathematically evaluated as

          $\epsilon = L * \frac{\Delta I}{ \Delta t }$          

=>         $\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }$

substituting values

             $\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }$  

             $\epsilon = 0.863 \ V$

The direction of the induced emf is  from a to b because according to Lenz's law the induced emf moves in the same direction as the current