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An air-filled toroidal solenoid has 300turns of wire, a mean radius of 12cm, and a cross-sectional area of 4cm2. If the current is 5A, calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

Short Answer

Expert verified

(a)B=2.5×10-3T

(b)L=6×10-5H

(c)U=7.5×10-4J

(d)u=2.49J/m3

(e)u=2.49J/m3

Step by step solution

01

The formulas used in the given problem

The magnetic field is given by B=μ0NI2πr.

The self-inductance of a solenoid is given by L=μ0N2A2πr.

The energy stored in a solenoid is given by U=12LI2.

The energy density is simply energy divided by the volume of the solenoid.

02

Calculate the magnetic field

Given thatr=12cm,A=5cm2,N=300andI=5A.

The magnetic field is given by

role="math" localid="1664190143190" B=μ0NI2πrB=4π×10-7×300×52π×0.12B=2.5×10-3T

03

Calculate the self-inductance and the energy stored

The self-inductance is given by

L=μ0N2A2πrL=4π×10-7×3002×4×10-42π×0.12L=6×10-5H

The energy stored is calculated by

U=12LI2U=12×6×10-5×52U=7.5×10-1J

04

Calculate the energy density by two ways

The energy density is given by

u=B22μ0u=(2.5×10-3)22×4π×10-7u=2.49J/m3

The energy density is also given by

u=EnergyVolumeu=U2πrAu=7.5×1042π×0.12×4×10-4u=2.49J/m3

So, by using both the formulas, we get the same value of energy density.

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