Question

A solenoid that is 66.1 cm long has a cross-sectional area of 13.8 cm2. There are 1410 turns of wire carrying a current of 8.01 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).

Answer 1

Answer:

a) Energy density of the magnetic field, u = 183.46 J/m³

b) Total energy, E = 0.167 J

Explanation:

a) Number of turns in the solenoid, N = 1410

Area, A = 13.8 cm² = 0.00138 m²

Current, I = 8.01 A

Length of the solenoid, l = 66.1 cm = 0.661 m

Energy density, u is given by the formula $u = \frac{B^2}{2 \mu_{0} }$

Where B is the magnetic field

The magnetic field in a solenoid is given by the formula, $B = \frac{N \mu_{0} I }{l}$

$B = \frac{1410 * 8.01* \mu_{0} }{0.661}$

$B = 17086.38 \mu_0$ T

$u = \frac{(17086.38 \mu_0)^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0}{2 }\\u = \frac{291944527.29 * 4\pi * 10^{-7} }{2 }\\u = 183.46 J/m^3$

b) The total energy = Energy density * Volume

E = u V

Volume = Area * Length

V = Al = 0.00138 * 0.661

V = 0.00091218 m³

E = 183.46 * 0.00091218

E = 0.167 J