The magnetic field along the axis of an air cored solenoid is B.The magnetic field energy density is?

Question

Answer

Magnetic Field Energy Density of an Air Cored Solenoid

The magnetic field energy density of an air cored solenoid can be calculated using the formula:

u = (B^2)/(2*mu0)

Explanation

The energy density of a magnetic field is the amount of energy that is stored per unit volume of the magnetic field. In the case of an air cored solenoid, the magnetic field is generated by the flow of current through the coil of wire. The strength of the magnetic field is measured in units of tesla (T).

The energy density of the magnetic field is determined by the magnetic field strength (B) and the permeability of free space (mu0), which is a constant value of 4*pi*10^-7 T*m/A. The formula for calculating the energy density is derived from the energy stored in the magnetic field:

U = (1/2) * L * I^2

where U is the energy stored in the magnetic field, L is the inductance of the solenoid, and I is the current flowing through the solenoid. The inductance of a solenoid is given by:

L = (mu0 * n^2 * A * l)/l

where n is the number of turns in the solenoid, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

Substituting the expression for inductance into the formula for energy stored in the magnetic field, and simplifying, we get:

U = (mu0 * n^2 * A * l * I^2)/(2*l)

Dividing the energy by the volume of the solenoid, we get the energy density:

u = U/V = (mu0 * n^2 * A * I^2)/(2*l)

Substituting the expression for magnetic field strength (B = mu0 * n * I) into the formula for energy density, we get:

u = (B^2)/(2*mu0)

This formula gives us the magnetic field energy density of an air cored solenoid in terms of the magnetic field strength.

Answer

Depend upon wire  wrapped on the solenoid.

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