Question

A solenoid of length 2.50 cm and radius 0.450 cm has 37 turns. If the wire of the solenoid has 1.85 amps of current, what is the magnitude of the magnetic field inside the solenoid? magnitude of the magnetic field: 0.00344 T Ignoring the weak magnetic field outside the solenoid, find the magnetic energy density inside the solenoid. magnetic energy density: J/m3 What is the total magnetic energy inside the solenoid? total magnetic energy:

Answer 1

Answer:

The total magnetic energy inside the solenoid is $1.872 *10^{-7} J$

Explanation:

$L = 0.025 m\\N = 37 turns\\I = 1.85 A\\r= 0.00450m\\B=0.00344T\\\mu = 4 \pi *10^{-7} TmA$

Magnetic field ,$B =\mu *N*I/L$

$B = \frac{\mu*1.85 * 37}{0.025}$

$B = 3.44*10^{-3} T$

The magnetic field is $3.44*10^{-3} T$

B) The energy density is gived in the equation,

$\eta = \frac{energy}{volume} = \frac{1}{2}\frac{B^2}{\mu}$

$\eta = 0.5 *(3.44*10^-3)^2/(4 \pi *10^{-7})$

Energy density $= 4.7084 J/m^3$

c) For the energy, we have,

$\Omega = \eta * V$

Where,

$\Omega$ =Energy

$\Omega =   4.7084* (pi*0.00450^2) * 0.025^2$

$\Omega= 1.872 *10^{-7} J$