Elements of Electromagnetics
Elements of Electromagnetics
7th Edition
ISBN: 9780190698669
Author: Sadiku
Publisher: Oxford University Press
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Chapter MA, Problem 1.15MA

(i)

To determine

To obtain: The value of A.

(i)

Expert Solution
Check Mark

Answer to Problem 1.15MA

The value of A is 2yax+3zay+5az_.

Explanation of Solution

Definition used:

The vector differential operator is basically called the del operator (), which in cartesian coordinate, is formulated as =xax+yay+zaz.

Given:

The vectors are given as A=2xyax+3zyay+5zaz and B=sinxax+2yay+5yaz.

Obtain A as follows.

A=(xax+yay+zaz)(2xyax+3zyay+5zaz)=x(2xy)ax+y(3zy)ay+z(5z)az=2yax+3zay+5az

Therefore, the value of A is 2yax+3zay+5az_.

(ii)

To determine

To obtain: The value of ×A.

(ii)

Expert Solution
Check Mark

Answer to Problem 1.15MA

The value of ×A is 3yax2xaz_.

Explanation of Solution

The vectors are given as A=2xyax+3zyay+5zaz and B=sinxax+2yay+5yaz.

Obtain ×A as follows.

×A=|axayazxyz2xy3zy5z|=[y(5z)z(3zy)]ax[x(5z)z(2xy)]ay+[x(3zy)y(2xy)]az=[03y]ax[00]ay+[02x]az=3yax2xaz

Therefore, the value of ×A is 3yax2xaz_.

(iii)

To determine

To obtain: The value of ×A.

(iii)

Expert Solution
Check Mark

Answer to Problem 1.15MA

The value of ×A is 0_.

Explanation of Solution

From part (b), the value of ×A is 3yax2xaz.

Obtain ×A as follows.

×A=(xax+yay+zaz)(3yax2xaz)=x(3y)axz(2x)az=(0)ax(0)az=0

Therefore, the value of ×A is 0_.

(iv)

To determine

To obtain: The value of A×B.

(iv)

Expert Solution
Check Mark

Answer to Problem 1.15MA

The value of ×A is 3yax2xaz_.

Explanation of Solution

The vectors are given as A=2xyax+3zyay+5zaz and B=sinxax+2yay+5yaz.

Obtain A×B as follows.

A×B=|axayaz2xy3zy5zsinx2y5y|=[3zy(5z)5z(2y)]ax[2xy(5z)5z(sinx)]ay+[2xy(2y)3zy(sinx)]az=[15yz210yz]ax[10xyz5zsinx]ay+[4xy23zysinx]az

Obtain A×B as follows.

A×B=(xax+yay+zaz)([15yz210yz]ax[10xyz5zsinx]ay+[4xy23zysinx]az)=x(15yz210yz)axy(10xyz5zsinx)ay+z(4xy23zysinx)az=0ax(10xz0)ay+(03ysinx)az=10xzay3ysinxaz

Therefore, the value of A×B is 10xzay3ysinxaz_.

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