Chapter 7, Problem 7.1P

1. Find H in rectangular components at P(2,3,4) if there is a current filament on the z axis carring 8 mA in the a_z direction. (b) Repeat if the filament is located at x=-1. Y=2. (c) Find H if both filaments are present.

To determine

(a)

The value of H for given rectangular component and current filament in the a_z direction.

Answer to Problem 7.1P

The value of H for given rectangular component and current filament in the a_z direction is $H_a=-294a_x+196a_y\text{μA/m}$

Explanation of Solution

Given:

Rectangular component at $P(2,3,4)$.

Current filament on z-axis carrying $8\text{mA}$.

Calculation:

Now we can apply Biot-Savart Law, we get;

   $\begin{array}{l}{H}_a={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{IdL\times a_R}{4\pi R^2}}\\ H_a={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[2a_x+3a_y+\left(4-z\right)a_z\right]}{4\pi {\left(z^2-8z+29\right)}^{\frac{3}{2}}}}\\ H_a={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[2a_y-3a_x\right]}{4\pi {\left(z^2-8z+29\right)}^{\frac{3}{2}}}}\end{array}$

Now using the integral table, this equation evaluates as

   $\begin{array}{l}{H}_a=\frac{I}{4\pi }{\left[\frac{2\left(2z-8\right)\left(2a_y-3a_x\right)}{52{\left(z^2-8z+29\right)}^{1/2}}\right]}^{\infty }{}_{-\infty }\\ H_a=\frac{I}{26\pi }\left(2a_y-3a_x\right)\\ I=8\text{mA}\\ H_a=-294a_x+196a_y\text{μA/m}\end{array}$

To determine

(b)

The value of H for given rectangular component and current filament, if it is located at $x=-1,y=2$.

Answer to Problem 7.1P

The value of H for given rectangular component and current filament is $H_b=-127a_x+382a_y\text{μA/m}$

Explanation of Solution

Given:

Current filament carrying $8\text{mA}$.

Current filament located at $x=-1,y=2$

Calculation:

Now we can apply Biot-Savart Law, we get;

   $\begin{array}{l}{H}_b={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{IdL\times a_R}{4\pi R^2}}\\ H_b={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[\left(2+1\right)a_x+\left(3-2\right)a_y+\left(4-z\right)a_z\right]}{4\pi {\left(z^2-8z+26\right)}^{\frac{3}{2}}}}\\ H_b={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[3a_y-a_x\right]}{4\pi {\left(z^2-8z+26\right)}^{\frac{3}{2}}}}\end{array}$

Now using the integral table, this equation evaluates as

   $\begin{array}{l}I=8\text{mA}\\ H_b=\frac{I}{4\pi }{\left[\frac{2\left(2z-8\right)\left(3a_y-a_x\right)}{40{\left(z^2-8z+26\right)}^{1/2}}\right]}^{\infty }{}_{-\infty }\\ H_b=\frac{I}{20\pi }\left(3a_y-a_x\right)\\ H_b=-127a_x+382a_y\text{μA/m}\end{array}$

To determine

(c)

The value of H if both filaments are present.

Answer to Problem 7.1P

The value of H if both filaments are present is $H_T=-421a_x+578a_y\text{μA/m}$

Explanation of Solution

Given:

Rectangular component at $P(2,3,4)$

Current filament located at carrying $x=-1,y=2$

Calculation:

For calculating $H_T$ when both filaments are present, so we just add the result of $H_a\text{and}{H}_b$.

For first filament:

Now we can apply Biot-Savart Law, we get;

   $\begin{array}{l}{H}_a={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{IdL\times a_R}{4\pi R^2}}\\ H_a={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[2a_x+3a_y+\left(4-z\right)a_z\right]}{4\pi {\left(z^2-8z+29\right)}^{\frac{3}{2}}}}\\ H_a={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[2a_y-3a_x\right]}{4\pi {\left(z^2-8z+29\right)}^{\frac{3}{2}}}}\end{array}$

Now using the integral table, this equation evaluates as

   $\begin{array}{l}{H}_a=\frac{I}{4\pi }{\left[\frac{2\left(2z-8\right)\left(2a_y-3a_x\right)}{52{\left(z^2-8z+29\right)}^{1/2}}\right]}^{\infty }{}_{-\infty }\\ H_a=\frac{I}{26\pi }\left(2a_y-3a_x\right)\\ I=8\text{mA}\\ H_a=-294a_x+196a_y\text{μA/m}\end{array}$

For second filament:

Now we can apply Biot-Savart Law, we get,

   $\begin{array}{l}{H}_b={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{IdL\times a_R}{4\pi R^2}}\\ H_b={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[\left(2+1\right)a_x+\left(3-2\right)a_y+\left(4-z\right)a_z\right]}{4\pi {\left(z^2-8z+26\right)}^{\frac{3}{2}}}}\\ H_b={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{Idz{a}_z\times \left[3a_y-a_x\right]}{4\pi {\left(z^2-8z+26\right)}^{\frac{3}{2}}}}\end{array}$

Now using the integral table, this equation evaluates as

   $\begin{array}{l}I=8\text{mA}\\ H_b=\frac{I}{4\pi }{\left[\frac{2\left(2z-8\right)\left(3a_y-a_x\right)}{40{\left(z^2-8z+26\right)}^{1/2}}\right]}^{\infty }{}_{-\infty }\\ H_b=\frac{I}{20\pi }\left(3a_y-a_x\right)\\ H_b=-127a_x+382a_y\text{μA/m}\end{array}$

   $\begin{array}{l}{H}_T=H_a+H_b\\ H_T=-421a_x+578a_y\text{μA/m}\end{array}$ $$