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

Answer 2

A) The magnetic field in the solenoid is : 0.1842 T

B) The energy density in the magnetic field if the solenoid if filled with air is :  1.3507 * 10⁻⁴  J/m³

C) The Total energy contained in the coil's magnetic field is : 0.2188 J

D) The inductance of the solenoid : 5.40 * 10⁻⁵ H

Given data:

Length of solenoid = 27 cm = 0.27 m

Cross-sectional area = 0.6 cm² = 0.6 * 10⁻⁴m²

Number of turns ( N ) = 440

Current in the solenoid ( i ) = 90 A

A) Determine the magnetic field in the solenoid

using the equation below

B = μo*N*i / L

   = ( 4π * 10⁻⁷ * 440 * 90 ) / 0.27

   = 0.1842 T

B) Calculate the energy density in the magnetic field

energy density = B² / 2 μo

                         = (0.1842)² /  ( 2 * 4π * 10⁻⁷ )

                         =  1.3507 * 10⁻⁴ J/m³

C) Calculate the total energy contained in the coil's magnetic field

Total energy = energy density * volume

                     = 1.3507 * 10⁴ * 0.27 * 0.6 * 10⁻⁴

                     = 0.2188 J

D) Calculate the inductance of the solenoid

Inductance ( L ) = 2U / i²

                          = ( 2 * 0.2188 ) / 90²

                          = 5.40 * 10⁻⁵ H

Hence we can conclude that the answers to your questions are as listed above.

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