A permanent magnet has
circular pole pieces of radius a separated by an arbitrary distance. The
magnetic field **B** is uniform between the pole pieces. Calculate the force
between the pole pieces in terms of B and a.

Solution:

- Concepts:

Energy stored in the magnetic field - Reasoning:

F = -dU/dx. - Details of the calculation:

The magnetic field is confined to the region between the pole pieces. The energy stored in the field is U = (2μ_{0})^{-½}∫_{all space}B^{2}dV in SI units. Here U = (2μ_{0})^{-½}B^{2}πa^{2}x, where x denotes the distance between the pole pieces. F = -dU/dx = -(2μ_{0})^{-½}B^{2}πa^{2}. The minus sign indicates that the force is attractive.

A solenoid is designed to store U_{L} = 0.10 J of energy when it carries
a current of I = 450 mA.

The solenoid has a cross-sectional area of A = 5.0 cm^{2}
and a length l = 0.20m.

How many turns of wire must the solenoid have?

Solution:

- Concepts:

The magnetic field inside a solenoid, magnetostatic energy - Reasoning

Approximation: B = μ_{0}nI inside the solenoid, B = 0 outside.

U = (2μ_{0})^{-½}∫_{all space}B^{2}dV.

The integral over all space reduces to an integral over the volume of the solenoid. - Details of the
calculation:

U_{L}= (1/(2μ_{0}))B^{2}V = μ_{0}n^{2}I^{2}lA/2.

2U_{L}/( μ_{0}I^{2}lA) = n^{2}, n = 88654. N = nl = 17731.

or

U_{L}= ½LI^{2}, L = nlBA/I = μ_{0}n^{2}lA, n^{2}= 2U_{L}/( μ_{0}I^{2}lA).

A spiral spring has N turns and initial length x_{0}. How does its length changes if a small current I is made to flow through it?
Assume that the spring has a spring constant k for longitudinal deformations,
that it can be treated as a perfect ("infinitely long") solenoid, and that its
radius R remains fixed.

Solution:

- Concepts:

In equilibrium F_{net}= 0. - Reasoning:

Adjacent coils of the spring attract each other. The spring gets shorter. As the spring contracts, the energy stored in the magnetic field changes and the battery maintaining the current I does work.

The mechanical work done by the system is ΔW_{mech}= ΔW_{batt}- ΔU_{mag}. In equilibrium we need F_{net}= 0, or dW_{mech}/dx = F_{ext}= kΔx. - Details of the calculation:

Given: I is small, Δx is small, the length of the spring changes little from x_{0}.

Let the x-axis point upward.

Assume the current I is established in a spring of length x_{0}.

The magnetic field inside the spring is uniform and has magnitude B = μ_{0}IN/x_{0}.

The energy stored in the field is U_{mag}= B^{2}V/(2μ_{0}) = B^{2}πR^{2}x_{0}/(2μ_{0}) = μ_{0}I^{2}N^{2}πR^{2}/(2x_{0}).

Assume the length of the spring changes by a small amount Δx.

For a small displacement the energy stored in the field changes by

ΔU = dU/dx|_{x0}Δx = -[μ_{0}I^{2}N^{2}πR^{2}/(2x_{0}^{2})]Δx.

The energy stored in the field increases when x decreases.

[We know the coil will contract, but the stored field energy increases when the coil contracts. How can this be?]

The flux through the coils is NBπR^{2}. When x changes, B changes, and therefore the flux changes and an emf is induced.

In the small time interval Δt we have emf = -NπR^{2}ΔB/Δt.

The extra work done by the battery to keep the current I flowing is emf*I*Δt = I*NπR^{2}ΔB.

For a small displacement ΔB = dB/dx|_{x0}Δx, so ΔW_{batt}= -[μ_{0}I^{2}N^{2}πR^{2}/x_{0}^{2}]Δx = 2ΔU.

When x decreases the battery has to do extra work against the induced emf to keep the current constant at I. (Lenz's rule)

But when x decreases, only ½ of the work the battery does ends up being stored as field energy. The other half goes into mechanical work done on the system, which can be stored as elastic potential energy or converted to kinetic energy, or dissipated through friction or radiation.

The mechanical work done by the system is ΔW_{mech}= ΔW_{batt}- ΔU_{mag}= -[μ_{0}I^{2}N^{2}πR^{2}/(2x_{0}^{2})]Δx.

When x decreases, ΔW_{mech}is positive.

When the system is in equilibrium if the net force is zero.

In equilibrium dW_{mech}/dx = -μ_{0}I^{2}N^{2}πR^{2}/(2x_{0}^{2}) = kΔx.

Δx = -μ_{0}I^{2}N^{2}πR^{2}/(2kx_{0}^{2}).

Why do we look at the force, and not just the energy?

ΔW_{mech}= -[μ_{0}I^{2}N^{2}πR^{2}/(2x_{0}^{2})]Δx.

When the coil has contracted by Δx and is in equilibrium after any oscillation have damped out, not all the mechanical work done is stored as elastic potential energy.

In this example F_{ext}= dW_{mech}/dx is constant.

The work done on the system by the external force to is W_{mech}= ∫_{0}^{Δx}F_{ext}dx = F_{ext}Δx.

(With our orientation of the x-axis, both F_{ext}and Δx are negative, W_{mech}is positive.)

But F_{ext}= kΔx, so W_{mech}= kΔx^{2}. The elastic potential energy stored in the spring is ½kΔx^{2}= ½W_{mech}.