Chapter 12

1.41 meters

$\frac{{\mu }_{0}I}{2R}$

4 amps flowing out of the page

Both have a force per unit length of $9.23\times {10}^{\text{−12}}\text{N/m}$

0.608 meters

In these cases the integrals around the Ampèrian loop are very difficult because there is no symmetry, so this method would not be useful.

a. 1.00382; b. 1.00015

a. $1.0\times {10}^{\mathrm{-4}}\text{T}$; b. 0.60 T; c. $6.0\times {10}^3$

Biot-Savart law’s advantage is that it works with any magnetic field produced by a current loop. The disadvantage is that it can take a long time.

If you were to go to the start of a line segment and calculate the angle $\theta$ to be approximately $0\text{°}$, the wire can be considered infinite. This judgment is based also on the precision you need in the result.

You would make sure the currents flow perpendicular to one another.

A magnetic field line gives the direction of the magnetic field at any point in space. The density of magnetic field lines indicates the strength of the magnetic field.

The spring reduces in length since each coil will have a north pole-produced magnetic field next to a south pole of the next coil.

Ampère’s law is valid for all closed paths, but it is not useful for calculating fields when the magnetic field produced lacks symmetry that can be exploited by a suitable choice of path.

If there is no current inside the loop, there is no magnetic field (see Ampère’s law). Outside the pipe, there may be an enclosed current through the copper pipe, so the magnetic field may not be zero outside the pipe.

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.

$5.66\times {10}^{\mathrm{-5}}\text{T}$

$B=\frac{{\mu }_{o}I}{8}\left(\frac{1}{a}-\frac{1}{b}\right)$ out of the page

$a=\frac{2R}{\pi }$; the current in the wire to the right must flow up the page.

20 A

Both answers have the magnitude of magnetic field of $4.5\times {10}^{\text{−5}}\text{T}.$

At P1, the net magnetic field is zero. At P2, $B=\frac{3{\mu }_{o}I}{8\pi a}$ into the page.

The magnetic field is at a minimum at distance a from the top wire, or half-way between the wires.

a. $F\text{/}l=2\times {10}^{\text{−5}}\text{N/m}$ away from the other wire; b. $F\text{/}l=2\times {10}^{\text{−5}}\text{N/m}$ toward the other wire

$B=\frac{{\mu }_{o}I}{2\pi a^{2}b}\left(\right(a_2+b_2)\widehat{i}+b\sqrt{(a^2–b^2)}\widehat{j})$

0.019 m

$\text{N}\times 6.28\times {10}^{\text{−5}}\text{T}$

$B=\frac{{\mu }_{o}I{R}^{2}N}{{\left({\left(\frac{d}{2}\right)}^2+R^2\right)}^{3\text{/}2}}$

a. ${\mu }_{0}I;$ b. 0; c. ${\mu }_{0}I;$ d. 0

a. $3{\mu }_{0}I;$ b. 0; c. $7{\mu }_{0}I;$ d. $\text{−2}{\mu }_{0}I$

at the radius R

$B=1.3\times {10}^{\mathrm{-2}}\text{T}$

roughly eight turns per cm

$B=\frac{1}{2}{\mu }_{0}nI$

0.0181 A

0.0008 T

317.31

$2.1\times {10}^{\mathrm{-4}}\text{A}·{\text{m}}^2$
$2.7\text{A}$

0.18 T

$B=1.4\times {10}^{\mathrm{-4}}\text{T}$

$3.2\times {10}^{\text{−19}}N$ in an arc away from the wire

a. above and below $B={\mu }_{0}j,$ in the middle $B=0;$ b. above and below $B=0,$ in the middle $B={\mu }_{0}j$

$\frac{dB}{B}=-\frac{dr}{r}$

a. 5026 turns; b. 0.00957 T

$B_1\left(x\right)=\frac{{\mu }_{0}I{R}^2}{2{\left(R^2+z^2\right)}^{3\text{/}2}}$

$B=\frac{{\mu }_0\sigma \omega }{2}R$

derivation

derivation

As the radial distance goes to infinity, the magnetic fields of each of these formulae go to zero.

a. $B=\frac{{\mu }_{0}I}{2\pi r}$; b. $B=\frac{{\mu }_0{J}_0{r}^2}{3R}$

$B(r)={\mu }_{0}NI\text{/}2\pi r$

$B=\frac{{\mu }_{0}I}{2\pi a}\text{ln}\frac{x+a}{x}.$

a. $B=\frac{{\mu }_0\sigma \omega }{2}\left[\frac{2h^2+R^2}{\sqrt{R^2+h^2}}\text{−2}h\right]$; b. $B=4.09\times {10}^{\mathrm{-5}}\text{T},$ 82% of Earth’s magnetic field