91Ó°ÊÓ

we need currents are needed to generate the magnetic field strengths of Table $29.1$at a point$1.0cm$ from a long, straight wire.

A current-carrying wire produces a magnetic field and the Biot-savart law enables us to calculate the magnitude and direction of this magnetic field at any point where the magnetic field due to the segment$∆\stackrel{→}{s}$of current-carrying wire is given by equation$\left(29.6\right)$

role="math" localid="1649184345177" ${\stackrel{→}{B}}_{currentsegment}=\frac{{\mathrm{μ}}_{\mathrm{ο}}}{4\mathrm{π}}\frac{I∆\stackrel{→}{s}×\hat{r}}{r^2}$

The infinitesimal wire segment $∆\stackrel{→}{s}$is in the same direction as the current I, r is the distance from $∆\stackrel{→}{s}$to the point and $\hat{r}$is a unit vector that points from $\stackrel{→}{dt}$to the point.

This approximation in equation $\left(1\right)$is only good if the length of the line segment is very small compared to the distance from the current element to the point .As the segment is very small compared with distance to the center of the loop we will use the approximation from.

The direction of $\stackrel{→}{dB}$is determined by applying the right-hand rule to the vector product $\stackrel{→}{ds}×\hat{r}$. Hence,the magnitude of $\stackrel{â‡\P Ä}{dB}$is given equation$\left(29.7\right)$in the form

$B=\frac{{\mathrm{μ}}_{\mathrm{ο}}}{2\mathrm{π}}\frac{I}{r}$

Our target is to find the current, so we solve equation $\left(2\right)$for I to be in the from

$I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}}$

From table $29.1,$the magnetic field at the surface of the earth is $B=5×{10}^{-5}T,$so the current that causes this magnetic field at distance $r=1cm=0.01m$is

$$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(5×{10}^{-5}T\right)\left(0.01m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =2.5A \end{gathered}$$

The magnetic field of the refrigerator magnet is $B=5×{10}^{-3}T$, so the current that causes this magnetic field at distance $r=1cm=0.01m$is

$$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(5×{10}^{-3}T\right)\left(0.01m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =250A \end{gathered}$$

The magnetic field of the laboratory magnet is $B=0.1$to $1$T, so the current that causes this range of the magnetic field at distance$r=1cm=0.01m$is

$$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(0.1T\right)\left(0.01m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =5000A \end{gathered}$$

to

$$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(0.01T\right)\left(0.01m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =50000A \end{gathered}$$

The magnetic of the superconducting magnet is$B=10T$, so current this magnetic field at distance $r=1cm=0.01m$is

$$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(10T\right)\left(0.01m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =5×{10}^{5}A \end{gathered}$$