Sawyer Burch

2023-03-24

The current density in a conducting medium is given by $\overrightarrow{J}=\frac{40}{{p}^{2}+1}\hat{{a}_{z}}mA/{m}^{2}$

The net current crossing the circular disc (0<p<5m) in z = 2 plane is ____ mA. (Answer upto 2 decimal places)

The net current crossing the circular disc (0<p<5m) in z = 2 plane is ____ mA. (Answer upto 2 decimal places)

rangiranitlh9

Beginner2023-03-25Added 6 answers

Step 1

Given: $I={\int}_{S}\overrightarrow{J}.\overrightarrow{ds}={\int}_{\varphi =0}^{2\pi}{\int}_{0}^{5}\frac{40}{{p}^{2}+1}pdpd\varphi $ mA

$=80\pi {\int}_{0}^{5}\frac{p}{{p}^{2}+1}d\rho mA$

$=\frac{80\pi}{2}\mathrm{ln}({p}^{2}+1){|}_{0}^{5}mA$

$=40\pi [\mathrm{ln}(26)-\mathrm{ln}(1)]$

Step 2

$=40\pi \times 3.258=409.42mA$

Given: $I={\int}_{S}\overrightarrow{J}.\overrightarrow{ds}={\int}_{\varphi =0}^{2\pi}{\int}_{0}^{5}\frac{40}{{p}^{2}+1}pdpd\varphi $ mA

$=80\pi {\int}_{0}^{5}\frac{p}{{p}^{2}+1}d\rho mA$

$=\frac{80\pi}{2}\mathrm{ln}({p}^{2}+1){|}_{0}^{5}mA$

$=40\pi [\mathrm{ln}(26)-\mathrm{ln}(1)]$

Step 2

$=40\pi \times 3.258=409.42mA$

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