Q25P - Introduction to Electrodynamics

Question

(a) Check product rule (iv) (by calculating each term separately) for the functions

 $\mathit{A}\mathbf{=}\mathit{x}\hat{\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{y}\hat{\mathbf{y}}\mathbf{+}\mathbf{3}\mathit{z}\mathbf{ }\hat{\mathbf{z}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{B}\mathbf{=}\mathbf{3}\mathit{x}\hat{\mathbf{x}}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{ }\hat{\mathbf{y}}$                         

(b) Do the same for product rule (ii).

(c) Do the same for rule (vi).

Step-by-Step Solution

Answer

a)   The product rule (iv) is proven.

b)   The product rule (ii) is proven.

c)   The product rule (vi) is proven.

Step 1: prove rule (iv)

To compute an expression substitute the vectors and other required expression and then simplify.

 

The vector A and B are defined as A=xi+2yj and B=3yi-2xj . Find th cross product of vectors A and B .

$$\begin{gathered} \mathbf{A}\mathbf{\times }\mathbf{B}\mathbf{=}\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \mathrm{x}& 2\mathrm{y}& 3\mathrm{z}\\ 3\mathrm{y}& -2\mathrm{x}& 0\end{array}\right| \\ =\left(0+6\mathrm{xz}\right)\mathrm{i}=(0+9\mathrm{yz})\mathrm{j}+(-2{\mathrm{x}}^2-6{\mathrm{y}}^2)\mathrm{k} \\ =6\mathrm{xzi}-9\mathrm{yzj}-2\left({\mathrm{x}}^2+3{\mathrm{y}}^2\right)\mathrm{k} ........ \left(1\right) \end{gathered}$$

The product rule (iv) is expressed as $\nabla .\left(A\times B\right)=B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$   

Compute left side of product rule (iv), as:

$$\begin{gathered} \nabla .\left(A\times B\right)=\left(\frac{\partial }{\partial X}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(6xzi-\left(9yzj9yzj\right)-2\left(x^2+3y^2\right)k\right) \\ =\frac{\partial }{\partial X}\left(6xzi\right)-\frac{\partial }{\partial y}\left(9yzj\right)-\frac{\partial }{\partial z}2\left(x^2+3y^2\right) \\ =9z+6z \\ =15z \end{gathered}$$

  Find the curl of vector A as follows: 

$$\begin{gathered} \nabla \times A=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& 2y& 3z\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(3z\right)-\frac{\partial }{\partial z}\left(2y\right)\right)i-\left(\frac{\partial }{\partial x}\left(3z\right)-\frac{\partial }{\partial z}\left(x\right)\right)j+\left(\frac{\partial }{\partial z}\left(3z\right)-\frac{\partial }{\partial y}\left(x\right)\right)k \\ =i\left(0-0\right)-j\left(0-0\right)+k\left(0-0\right) \\ =0 \end{gathered}$$

The value of $\nabla \times A$  is 0,, then value of $B.\left(\nabla \times A\right)$  is also zero.

 

Find the curl of vector $B$ as follows:

$$\begin{gathered} \nabla \times B=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 3y& -2x& 0\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(0\right)-\frac{\partial }{\partial z}\left(-2x\right)\right)i-\left(\frac{\partial }{\partial x}\left(0\right)-\frac{\partial }{\partial z}\left(-3y\right)\right)j+\left(\frac{\partial }{\partial x}\left(-2x\right)-\frac{\partial }{\partial y}\left(3y\right)\right)k \\ =-5k \end{gathered}$$

Now compute the value of $A.\left(\nabla \times B\right)$ , as follows:

 

 $$\begin{gathered} A.\left(\nabla \times B\right)=\left(xi+2yj+3zk\right).\left(-5k\right) \\ =-15k \end{gathered}$$

 

compute the value of  $B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$, as follows 

 

 $$\begin{gathered} B.\left(\nabla \times A\right)-A\left(\nabla \times B\right)=0-\left(-15k\right) \\ =15k \end{gathered}$$

 

This concludes that the value of $\nabla .\left(A\times B\right)$ is equal to that of  

$B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$. Thus $\nabla .\left(A\times B\right)=B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$ . Hence rule (iv) is proven.

Step 2: prove product rule (ii)

To compute an expression substitute the vectors and other required expression and then simplify

The product rule (ii) is expressed as

$\nabla \left(A.B\right)A\times \left(\nabla \times B\right)-B\times \left(\nabla \times A\right)+\left(A.\nabla \right)B+\left(B.\nabla \right)A$

Compute left side of product rule (ii), as:

$$\begin{gathered} \nabla \left(A.B\right)=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(xi+2yj+3zk\right).\left(3yj-2xj\right) \\ =\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(3xy-4xy\right) \\ =\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(-xy\right) \\ =i\frac{\partial }{\partial x}\left(-xy\right)+j\frac{\partial }{\partial y}\left(-xy\right)+k\frac{\partial }{\partial z}\left(-xy\right) \\ =-yi-xj \end{gathered}$$ 

It is already computed that $\nabla \times A$  is 0 , then value of$B\times \left(\nabla \times A\right)$is also zero and$\nabla \times B is -5k$ .

 

Find the value of $A\times \left(\nabla \times B\right)$ as follows:

$$\begin{gathered} A\times \left(\nabla \times B\right)=\left|\begin{array}{ccc}i& j& k\\ x& 2y& 3z\\ 0& 0& -5\end{array}\right| \\ =i\left(-10y-0\right)-j\left(-5x-0\right)+z\left(0\right) \\ =-10\mathrm{yi}+5\mathrm{xj} \\ \mathrm{Find} \mathrm{the} \mathrm{curl} \mathrm{of} \mathrm{vector} \mathrm{B} \mathrm{as} \mathrm{follows}: \\ \nabla \times \mathrm{B}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \frac{\partial }{\partial \mathrm{x}}& \frac{\partial }{\partial \mathrm{y}}& \frac{\partial }{\partial \mathrm{z}}\\ 3\mathrm{y}& -2\mathrm{x}& 0\end{array}\right| \\ =\left(\frac{\partial }{\partial \mathrm{y}}\left(0\right)-\frac{\partial }{\partial \mathrm{z}}\left(-2\mathrm{x}\right)\right)\mathrm{i}-\left(\frac{\partial }{\partial \mathrm{y}}\left(0\right)-\frac{\partial }{\partial \mathrm{z}}\left(3\mathrm{y}\right)\right)\mathrm{j}+\left(\frac{\partial }{\partial \mathrm{x}}\left(-2\mathrm{x}\right)-\frac{\partial }{\partial \mathrm{y}}\left(3\mathrm{y}\right)\right)\mathrm{k} \\ =-5\mathrm{k} \end{gathered}$$ 

Find the value of  $\left(A.\nabla \right)B$.

$$\begin{gathered} \left(A.\nabla \right)B=\left(x\frac{\partial }{\partial x}+2y\frac{\partial }{\partial y}+3z\frac{\partial }{\partial z}\right)\left(3yi-2xj\right) \\ =x\frac{\partial }{\partial x}\left(3yi-2xj\right)+2y\frac{\partial }{\partial y}\left(3yi-2xj\right)+3z\frac{\partial }{\partial z}\left(3yi-2xj\right) \\ =x\left(-2\right)i=2y\left(3\right)j+3z\left(0\right)k \\ =-2xi+6yj \end{gathered}$$

Find the value of  $\left(B.\nabla \right)A$.

$$\begin{gathered} \left(B.\nabla \right)A=\left(3y\frac{\partial }{\partial x}-2x\frac{\partial }{\partial y}\right)\left(xi-2yj+3zk\right) \\ =3y\frac{\partial }{\partial x}\left(xi-2xj+3zk\right)-2x\frac{\partial }{\partial y}\left(xi-2yj+3zk\right) \\ =3yi-2x\left(2\right)j \\ =3yi-4xj \end{gathered}$$

Step 3: compute right side of rule (ii)

To compute an expression substitute the vectors and other required expression and then simplify.

 

$$\begin{gathered} \mathrm{Substitiute}3yi-4xj \mathrm{for} \left(\mathrm{B}.\nabla \right)\mathrm{A} ,-2\mathrm{xi}+6\mathrm{yj} \mathrm{for} \left(\mathrm{A}.\nabla \right)\mathrm{B}, \mathrm{for} \mathrm{B}\times \left(\nabla .\mathrm{A}\right) \\ \mathrm{and} -10\mathrm{yi}+5\mathrm{xj} \mathrm{for} \mathrm{A}\times \left(\nabla \times \mathrm{B}\right) \mathrm{into} \end{gathered}$$   

 

$$\begin{gathered} \mathrm{Thus} \mathrm{the} \mathrm{value} \mathrm{of} \nabla \left(\mathrm{A}.\mathrm{B}\right) \mathrm{is} \mathrm{same} \mathrm{as} \mathrm{that} \mathrm{of} \\ \mathrm{A}\times \left(\nabla \times \mathrm{B}\right)-\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)+\left(\mathrm{A}.\nabla \right)\mathrm{B}+\left(\mathrm{B}.\nabla \right)\mathrm{A}. \mathrm{Hence} \mathrm{the} \mathrm{rule} \left(\mathrm{ii}\right) \mathrm{is} \mathrm{proven}. \end{gathered}$$

Step 4: Prove product rule (vi)

To compute an expression substitute the vectors and other required expression and then simplify

 

The product rule (iv) is expressed as  

$\nabla \times \left(A\times B\right)=\left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)$

 

Compute left side of product rule (iv), as:

$$\begin{gathered} \nabla \times \left(A\times B\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& 2y& 3z\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(3z\right)-\frac{\partial }{\partial z}\left(2y\right)\right)i-\left(\frac{\partial }{\partial x}\left(3z\right)-\frac{\partial }{\partial z}\left(x\right)\right)j+\left(\frac{\partial }{\partial z}\left(2y\right)-\frac{\partial }{\partial y}\left(x\right)\right)k \\ =i\left(-12y-19z\right)-j\left(-4x-6x\right)+k\left(0-0\right) \\ =-21yi+10xj \end{gathered}$$ 

Find the value of $B\left(\nabla .A\right)$ .

$$\begin{gathered} B\left(\nabla .A\right)=\left(3yi-2xj\right)\left(\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(2y\right)+\frac{\partial }{\partial z}\left(3z\right)\right) \\ =\left(3yi-2xj\right)\left(1+2+3\right) \\ =6\left(3yi-2xj\right) \\ =18yi-12xj \end{gathered}$$

Find the value of $\left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+\left(\nabla .A\right)A\left(\nabla .B\right)$ .

 

 $$\begin{gathered} \left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)=\left[\begin{array}{c}\left(3yi-4xj\right)-\left(-2xj+6yi\right)\\ +0-\left(18yi-12xj\right)\end{array}\right] \\ =-21yi+10xj \end{gathered}$$$$\begin{gathered} \left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)=\left[\begin{array}{c}\left(3y-4xj\right)\left(-2xj-6yi\right)\\ +0-\left(18yi-12xj\right)\end{array}\right] \\ =-21yi+10xj \end{gathered}$$

 

Thus the value of $\nabla \times \left(A\times B\right)$  is same as that of

$\left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)$Hence the rule (iv) is proven.