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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. $1.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\text{T}$ ; b. 0.60 T; c. $6.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}$
Permeability of free space | ${\mu}_{0}=4\pi \phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\text{T}\cdot \text{m/A}$ |
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 | $\overrightarrow{B}=\frac{{\mu}_{0}}{4\pi}{\displaystyle \underset{\text{wire}}{\int}\frac{Id\overrightarrow{l}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\widehat{r}}{{r}^{2}}}$ |
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 | $\oint \overrightarrow{B}}\xb7d\overrightarrow{l}={\mu}_{0}I$ |
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 | $\mu =(1+\chi ){\mu}_{0}$ |
Magnetic field of a solenoid
filled with paramagnetic material |
$B=\mu nI$ |
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.
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 $2.1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-23}}\text{A}\xb7{\text{m}}^{2}.$ (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?
$2.1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\text{A}\xb7{\text{m}}^{2}$
$2.7\phantom{\rule{0.2em}{0ex}}\text{A}$
Suppose you wish to produce a 1.2-T magnetic field in a toroid with an iron core for which $\chi =4.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}.$ 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 $\chi =3.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3},$ what is the magnetic field within it?
0.18 T
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