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Sagot :
The question is incomplete. The complete question is :
A long, straight wire lies along the z-axis and carries a 3.90-A current in the + z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mm segment of the wire centered at the origin.
A) x=2.00m,y=0, z=0
Bx,By,Bz = ? T
Enter your answers numerically separated by commas.
B) x=0, y=2.00m, z=0
C) x=2.00m, y=2.00m, z=0
D) x=0, y=0, z=2.00m
Solution :
The expression of the magnetic field using the Biot Savart's law is given by :
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
a). The position vector is on the positive x direction.
[tex]$\vec r = (2 \ m) \ \hat i$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat i }{4 \pi \times (2)^3}$[/tex]
[tex]$d \vec B=(5.85 \times 10^{-11} \ T)\hat j$[/tex]
The magnetic field is [tex]$(0, \ 5.85 \times 10^{-11} \ T, \ 0).$[/tex]
b). The position vector is in the positive y-direction.
[tex]$\vec r = (2 \ m) \ \hat j$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat j }{4 \pi \times (2)^3}$[/tex]
[tex]$d \vec B=(5.85 \times 10^{-11} \ T)(-\hat{i})$[/tex]
The magnetic field is [tex]$(- 5.85 \times 10^{-11} \ T, \ 0, \ 0).$[/tex]
c). The position vector is :
[tex]$\vec r = (2)\hat i + (2)\hat j$[/tex]
[tex]$|\vec r| = \sqrt{(2)^2+(2)^2}$[/tex]
[tex]$=2.828 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times ((2)\hat i + (2) \hat j) }{4 \pi \times (2.828)^3}$[/tex]
[tex]$=(4.13\times 10^{-11})\hat j+(4.13\times 10^{-11})(-\hat i)$[/tex]
The magnitude of the magnetic field is :
[tex]$|d\vec B|=\sqrt{(4.13\times 10^{-11})^2+(4.13\times 10^{-11})^2}$[/tex]
[tex]$=5.84 \times 10^{-11} \ T$[/tex]
Therefore, the magnetic field is [tex]$(-4.13 \times 10^{-11}\ T, \ 4.13 \times 10^{-11}\ T, \ 0 )$[/tex]
d). The position vector is in the positive y-direction.
[tex]$\vec r = (2 \ m) \ \hat k$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat k }{4 \pi \times (2)^3}$[/tex]
= 0 T
The magnetic field is (0, 0, 0)
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