Question

Chapter 29, Problem 075 The gure shows a wire segment 0 length Δs-4.5 cm, centered at the origin, carrying current i-41 A n the posti e y direction as art of some complete Circuit To calculate the magnitude of the magn tic e produc 2 by the segment at a point several meters from the origin, we can use the Biot Sa art law as- μ 4m Δ s sin θ This is because rand sa e essentially constant o er the segment nunt ec or notation at al ulat the (x, y, z) coordinates (a) (0, 0, 8.0 m), (b) (0, 8.2 m, 0), (c) (8.6 m, 7.3 m, 0), and (d) (-5.3 m,-5.6 m,0) (a) 8 (o, o, B.0 m)(Number Units Byt (b) (O, 8.2 m, 0) Number Units (c) 8 (8.6 m, 7.3 m, 0)-(Number Units (d (-5.3 m,-5.6 m,0) - (Number Units Show Work is REQUIRED for this question: Open Show Wark

Answer 1

The vector form of the Biot Savart equation is the following:

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The first part of the right hand side of the equation is constant:

$\vec{B}=4,1\times 10^{-7}N/A(\frac{\vec{\Delta{S}}\times\vec{r}}{r^{3}})$

The length vector will also be constant:

(0: 0. 0225m : 0)-(0:-0. 0225m: 0)-(0: 0. 045m : 0) ▲S

a)

(0: 0: 8m

AS0 0,045m 0(0, 36m2)i 0 0 8m

$r^{3}=(\sqrt{0^{2}+0^{2}+(8m)^{2}})^{3}=512m^{3}$

, 36m 512m3 --(2.88 × 10-10T)

b)

= (0: 8.2771:0)

ASXT.=|0 0|=0 0.045m 0 8,2m 0

$\rightarrow B=0$

c)

$\vec{r}=(8,6m;7,3m;0)$

$\vec{\Delta S}\times \vec{r}=\begin{vmatrix} \vec{i }&\vec{j}&\vec{k} \\ 0&0,045m&0 \\ 8,6m&7,3m&0 \end{vmatrix}=(-0,387m^{2})\vec{k}$

$r^{3}=(\sqrt{(8,6m)^{2}+(7,3m)^{2}+0^{2}})^{3}=1435,445m^{3}$

(-0.387m2)k) = (-1. 105 × 10-10T)k 1435,445m3 4, 1

d)

$\vec{r}=(-5,3m;-5,6m;0)$

$\vec{\Delta S}\times \vec{r}=\begin{vmatrix} \vec{i }&\vec{j}&\vec{k} \\ 0&0,045m&0 \\ -5,3m&-5,6m&0 \end{vmatrix}=(0,2385m^{2})\vec{k}$

$r^{3}=(\sqrt{(-5,3m)^{2}+(-5,6m)^{2}+0^{2}})^{3}=458,38m^{3}$

$\vec{B}=4,1\times 10^{-7}N/A(\frac{(0,2385m^{2})\vec{k}}{458,38m^{3}})=(2,13\times 10^{-10}T)\vec{k}$