12.7 Magnetism in matter  (Page 5/13)

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Check Your Understanding Repeat the calculations from the previous example for ${I}_{0}=0.040\phantom{\rule{0.2em}{0ex}}\text{A}.$

a. $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{T}$ ; b. 0.60 T; c. $6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}$

Summary

• Materials are classified as paramagnetic, diamagnetic, or ferromagnetic, depending on how they behave in an applied magnetic field.
• Paramagnetic materials have partial alignment of their magnetic dipoles with an applied magnetic field. This is a positive magnetic susceptibility. Only a surface current remains, creating a solenoid-like magnetic field.
• Diamagnetic materials exhibit induced dipoles opposite to an applied magnetic field. This is a negative magnetic susceptibility.
• Ferromagnetic materials have groups of dipoles, called domains, which align with the applied magnetic field. However, when the field is removed, the ferromagnetic material remains magnetized, unlike paramagnetic materials. This magnetization of the material versus the applied field effect is called hysteresis.

Key equations

 Permeability of free space ${\mu }_{0}=4\pi \phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\text{T}\cdot \text{m/A}$ Contribution to magnetic field from a current element $dB=\frac{{\mu }_{0}}{4\pi }\phantom{\rule{0.2em}{0ex}}\frac{I\phantom{\rule{0.2em}{0ex}}dl\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta }{{r}^{2}}$ Biot–Savart law $\stackrel{\to }{B}=\frac{{\mu }_{0}}{4\pi }\underset{\text{wire}}{\int }\frac{Id\stackrel{\to }{l}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{^}{r}}{{r}^{2}}$ Magnetic field due to a long straight wire $B=\frac{{\mu }_{0}I}{2\pi R}$ Force between two parallel currents $\frac{F}{l}=\frac{{\mu }_{0}{I}_{1}{I}_{2}}{2\pi r}$ Magnetic field of a current loop $B=\frac{{\mu }_{0}I}{2R}\phantom{\rule{0.2em}{0ex}}\text{(at center of loop)}$ Ampère’s law $\oint \stackrel{\to }{B}·d\stackrel{\to }{l}={\mu }_{0}I$ Magnetic field strength inside a solenoid $B={\mu }_{0}nI$ Magnetic field strength inside a toroid $B=\frac{{\mu }_{o}NI}{2\pi r}$ Magnetic permeability $\mu =\left(1+\chi \right){\mu }_{0}$ Magnetic field of a solenoid filled with paramagnetic material $B=\mu nI$

Conceptual questions

A diamagnetic material is brought close to a permanent magnet. What happens to the material?

If you cut a bar magnet into two pieces, will you end up with one magnet with an isolated north pole and another magnet with an isolated south pole? Explain your answer.

The bar magnet will then become two magnets, each with their own north and south poles. There are no magnetic monopoles or single pole magnets.

Problems

The magnetic field in the core of an air-filled solenoid is 1.50 T. By how much will this magnetic field decrease if the air is pumped out of the core while the current is held constant?

A solenoid has a ferromagnetic core, n = 1000 turns per meter, and I = 5.0 A. If B inside the solenoid is 2.0 T, what is $\chi$ for the core material?

317.31

A 20-A current flows through a solenoid with 2000 turns per meter. What is the magnetic field inside the solenoid if its core is (a) a vacuum and (b) filled with liquid oxygen at 90 K?

The magnetic dipole moment of the iron atom is about $2.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\text{A}·{\text{m}}^{2}.$ (a) Calculate the maximum magnetic dipole moment of a domain consisting of ${10}^{19}$ iron atoms. (b) What current would have to flow through a single circular loop of wire of diameter 1.0 cm to produce this magnetic dipole moment?

$2.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{A}·{\text{m}}^{2}$
$2.7\phantom{\rule{0.2em}{0ex}}\text{A}$

Suppose you wish to produce a 1.2-T magnetic field in a toroid with an iron core for which $\chi =4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}.$ The toroid has a mean radius of 15 cm and is wound with 500 turns. What current is required?

A current of 1.5 A flows through the windings of a large, thin toroid with 200 turns per meter. If the toroid is filled with iron for which $\chi =3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3},$ what is the magnetic field within it?

0.18 T

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