Question

A solenoid 27.0 cm long and with a cross-sectional area of 0.600 cm2 contains 440 turns of wire and carries a current of 90.0 A. Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic field if the solenoid is filled with air; (c) the total energy contained in the coil’s magnetic field (assume the field is uniform); (d) the inductance of the solenoid

Answer 1

Answer:

(A) 0.1842 T (B) $1.3507\times 10^{-4}J/m^3$ (C) 0.2188 J (D) $5.40\times 10^{-5}H$

Explanation:

Length of the solenoid L = 27  cm =0.27 m

Area of cross section $A=0.6\ cm^2=0.6\times 10^{-4}m^2$

Number of turns N = 440

Current i = 90 A

(A) Magnetic field in the solenoid $B=\mu _0ni=\frac{\mu _0Ni}{l}=\frac{4\pi \times 10^{-7}\times 440\times 90}{.27}=0.1842\ T$

(B) The energy density is given by $energy\ density =\frac{B^2}{2\mu _0}=\frac{0.1842^2}{2\times 4\pi \times 10^{-7}}=1.3507\times 10^4\ j/m^3$

(C) The total energy contained in the coli magnetic field = energy density ×volume = energy density ×l×A

So the total energy $=1.3507\times 10^4\times 0.27\times 0.6\times 10^{-4}=0.2188\ j$

(D) The energy stored in the inductor is given by $U=\frac{1}{2}Li^2$

So $L=\frac{2U}{i^2}=\frac{2\times 0.2188}{90^2}=5.40 \times 10^{-5}H$