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# Magnetic Materials Help (page 2)

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By McGraw-Hill Professional
Updated on Sep 7, 2011

#### Problem

Suppose that a metal rod is surrounded by a coil and that the magnetic flux density can be made as great as 0.500 T; further increases in current cause no further increase in the flux density inside the core. Then the current is removed; the flux density drops to 500 G. What is the retentivity of this core material?

#### Solution

First, convert both flux density figures to the same units. Remember that 1 T = 10 4 G. Thus the flux density is 0.500×10 4 =5,000 G with the current and 500 G without the current. “Plugging in” these numbers gives us this:

B r = 100 × 500/5,000 = 100 × 0.100 = 10.0 percent

## Permanent Magnets

Any ferromagnetic material, or substance whose atoms can be aligned permanently, can be made into a permanent magnet. These are the magnets you played with as a child (and maybe still play with when you use them to stick notes to your refrigerator door). Some alloys can be made into stronger permanent magnets than others.

One alloy that is especially suited to making strong permanent magnets is known by the trade name Alnico . This word derives from the chemical symbols of the metals that comprise it: aluminum (Al), nickel (Ni), and cobalt (Co). Other elements are sometimes added, including copper and titanium. However, any piece of iron or steel can be magnetized to some extent. Many technicians use screwdrivers that are slightly magnetized so that they can hold onto screws when installing or removing them from hard-to-reach places.

Permanent magnets are best made from materials with high retentivity. They are made by using the material as the core of an electromagnet for an extended period of time. If you want to magnetize a screwdriver a little bit so that it will hold onto screws, stroke the shaft of the screwdriver with the end of a bar magnet several dozen times. However, take note: Once you have magnetized a tool, it is practically impossible to completely demagnetize it.

## Flux Density Inside A Long Coil

Suppose that you have a long coil of wire, commonly known as a solenoid , with n turns and whose length in meters is s . Suppose that this coil carries a direct current of I amperes and has a core whose permeability is μ. The flux density B in teslas inside the core, assuming that it is not in a state of saturation, can be found using this formula:

B = 4π × 10 −7nI/s )

A good approximation is

B = 1.2566 × 10 −6nI/s )

#### Problem

Consider a dc electromagnet that carries a certain current. It measures 20 cm long and has 100 turns of wire. The flux density in the core, which is known not to be in a state of saturation, is 20 G. The permeability of the core material is 100. What is the current in the wire?

#### Solution

As always, start by making sure that all units are correct for the formula that will be used. The length s is 20 cm, that is, 0.20 m. The flux density B is 20 G, which is 0.0020 T. Rearrange the preceding formula so it solves for I :

B = 1.2566 × 10 −6nI/s )

B/I = 1.2566 × 10 −6n/s )

I −1 = 1.2566 × 10 −6n/sB

I = 7.9580 × 10 5 ( sBn )

This is an exercise, but it is straightforward. Derivations such as this are subject to the constraint that we not divide by any quantity that can attain a value of zero in a practical situation. (This is not a problem here. We aren’t concerned with scenarios involving zero current, zero turns of wire, permeability of zero, or coils having zero length.) Let’s “plug in the numbers”:

I = 7.9580 × 10 5 (0.20 × 0.0020)/(100 × 100)

= 7.9580 × 10 5 × 4.0 × 10 −8

= 0.031832 A = 31.832 mA

This must be rounded off to 32 mA because we are only entitled to claim two significant figures.

Practice problems of these concepts can be found at: Magnetism Practice Test

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