Question

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
(a) find the current density at P(rho = 3, φ = 30◦, z = 2);
(b) determine the total current flowing outward through the circular band rho = 3, 0 < φ < 2π, 2 < z < 2.8

Answer 1

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)$.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point (P), we can simply replace the coordinates in the current density equation.  Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0<\phi<2\pi, \ 2<z<2.8$. Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

Answer 2

Following are the solution to the given points:

Calculating the current density:

 $\bold{J = 10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi} \frac{mA}{m^2}}{}$

For point a:

Calculating the current density at point P:  

$\to p=2 \\\\ \to \phi =60^{\circ}\\\\ \to Z=3$

$\to J= 10(2)^2 3 (3) a_{\rho} - (4\times 2 \times ( \cos 60^{\circ})^2) a_{\phi}$

       $=120 \ a_{\rho} - 4 \times 2 \times \frac{1}{2^2} a_{\phi} \\\\= 120 a_{\phi} - 2 a_{\phi} \frac{mA}{m^2}$

For point b:

Calculating the total current:  

$\to I=\int \int \vec{J} \cdot \vec{ds}$

       $= \int \int (10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi}) \cdot (\rho d \phi \cdot dz) a_{\rho}$

Note:  

$\text{p=constont then} \vec{ds} = \rho d \phi dz \hat{a}_{\rho}$

$I= \int \int (10 \rho^2 z) \cdot ( d \phi dz)$

  $= (10 \rho^3 \int^{8}_{3} z dz \int^{2 \pi}_{\pi} 1 \cdot d \phi\\\\= (10 \times 4^3 [\frac{z^2}{2}]^{8}_{3} \times [\phi]^{2 \pi}_{\pi}\\\\= 10 \times 64 \times \frac{(64-9)}{2} \times (2\pi-\pi) MA \\\\= 55292 \ MA \\\\= 55.292\ A$

Learn more:

brainly.com/question/14821891