Answer:
(a) The magnetic energy density in the field is 6.366 J/m³
(b) The energy stored in the magnetic field within the solenoid is 5 kJ
Explanation:
magnitude of magnetic field inside solenoid, B = 4 T
inner diameter of solenoid, d = 6.2 cm
inner radius of the solenoid, r = 3.1 cm = 0.031 m
length of solenoid, L = 26 cm = 0.26 m
(a) The magnetic energy density in the field is given by;
![u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3](https://tex.z-dn.net/?f=u%20_B%20%3D%20%5Cfrac%7BB%5E2%7D%7B2%5Cmu_o%7D%20%5C%5C%5C%5Cu%20_B%20%3D%20%5Cfrac%7B%284%29%5E2%7D%7B2%284%5Cpi%2A10%5E%7B-7%7D%29%7D%5C%5C%5C%5Cu_B%20%3D%206.366%2A10%5E6%20%5C%20J%2Fm%5E3)
(b) The energy stored in the magnetic field within the solenoid
![U_B = u_B V\\\\U_B = u_B AL](https://tex.z-dn.net/?f=U_B%20%3D%20u_B%20V%5C%5C%5C%5CU_B%20%3D%20u_B%20AL)
![U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\](https://tex.z-dn.net/?f=U_B%20%3D%20u_B%28A%29%28L%29%5C%5C%5C%5CU_B%20%3D%206.366%2A10%5E6%28%5Cpi%20%2A%200.031%5E2%29%280.26%29%20%5C%5C%5C%5CU_B%20%3D%204997.69%20J%5C%5C%5C%5CU_B%20%3D%205%20%5C%20KJ%5C%5C)