Question

A 299 turn solenoid has a radius of 4.90 cm and a length of 16.0 cm.

(a) Find the inductance of the solenoid.

(b) Find the energy stored in it when the current in its windings is 0.495 A.

Answer 1

A) To find he inductance, the eq is:

$L = \frac{\mu_0N^2A}{l}$

Where:

$\mu_0 = 4\pi \times 10^{-7} T\cdot m/A$

$A = \pi r^2 = (0.049m)^2 \times \pi = 7.543 \times 10^{-3}m^2$

Answer:

$L = \frac{\mu_0N^2A}{l} = \frac{(4\pi \times 10^{-7}T\cdot m/A)(299)^2(7.543\times 10^{-3})}{0.16m} = 5.296 \times 10^{-3}H$

b) The energy is:

$E = \frac{1}{2} LI^2 = (0.5)(5.296\times 10^{-3}H)(0.495A)^2 = 6.488\times 10^{-4}J$