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54, 5.11, 5.22, 5.37, 5.39 6.4, 6.36, 6.40, 6.60, 6.63 In a cylindrical conductor of radius 4mm, current density is $J = 5e^{-10} \hat{z}$ A/m$^2$ Find current through conductor $I = \int \vec{J} \cdot d\vec{s}$ $r = 4mm$ $\vec{J} = 5e^{-10} \hat{z} \rightarrow$ convert to sph to use r $A_r = 5e^{-10} \cos\theta$ $A_\theta = -5e^{-10} \sin\theta$ $J = 5e^{-10} \cos\theta \hat{r} - 5e^{-10} \sin\theta \hat{\theta}$ $d\vec{s} = $ $\int_0^{2\pi} \int_0^{\pi} (5e^{-10} \cos\theta \hat{r} - 5e^{-10} \sin\theta \hat{\theta}) d\vec{s}$ $r = 4mm

          54, 5.11, 5.22, 5.37, 5.39
6.4, 6.36, 6.40, 6.60, 6.63
In a cylindrical conductor of radius 4mm, current density is $J = 5e^{-10} \hat{z}$ A/m$^2$
Find current through conductor
$I = \int \vec{J} \cdot d\vec{s}$
$r = 4mm$
$\vec{J} = 5e^{-10} \hat{z} \rightarrow$ convert to sph to use r
$A_r = 5e^{-10} \cos\theta$
$A_\theta = -5e^{-10} \sin\theta$
$J = 5e^{-10} \cos\theta \hat{r} - 5e^{-10} \sin\theta \hat{\theta}$
$d\vec{s} = $
$\int_0^{2\pi} \int_0^{\pi} (5e^{-10} \cos\theta \hat{r} - 5e^{-10} \sin\theta \hat{\theta}) d\vec{s}$  $r = 4mm

54, 5.11, 5.22, 5.37, 5.39
6.4, 6.36, 6.40, 6.60, 6.63
In a cylindrical conductor of radius 4mm, current density is J = 5e^-10ẑ A/m^2
Find current through conductor
I = ∫J⃗· ds⃗
r = 4mm
J⃗ = 5e^-10ẑ→ convert to sph to use r
Ar = 5e^-10cosθ
Aθ = -5e^-10sinθ
J = 5e^-10cosθr̂ - 5e^-10sinθθ̂
ds⃗ =
∫0^2π∫0^π (5e^-10cosθr̂ - 5e^-10sinθθ̂) ds⃗ r = 4mm

Added by Abigail J.

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