I think the answer is 2283g
<span>On the y-axis (the bottom of the table) hope this helps</span>
Answer:
Explanation: This is done using the equation:

Because the Radius is a know value. We have the following.

Which is:
4188.7902 mm
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
52800000000000000000000000000000000000000000