Chapter 8, Problem 13P

(a)

To determine

Find the force on side 1 of the triangular loop.

Answer to Problem 13P

The force on side 1 of the triangular loop is $2.197a_z\mu \text{N}$.

Explanation of Solution

Calculation:

Write the general expression to calculate the force.

$F={\displaystyle {\oint }_{L}Idl\times B}$        (1)

Here,

$Idl$ is the current element, and

$B$ is the magnetic flux density.

Refer to the mentioned Figure given in the textbook.

Using equation (1), the force on side 1 of the triangular loop is calculated as follows with integration limits along $\rho$-direction.

$\begin{array}{c}{F}_1=I_1{\displaystyle \int d\rho a_{\rho }\times \left(\frac{{\mu }_{\text{o}}{I}_2}{2\pi \rho }\right)a_{\varphi }}\left\{\because dl=d\rho a_{\rho },B=\frac{{\mu }_{\text{o}}{I}_2}{2\pi \rho }{a}_{\varphi }\right\}\\ =\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }{\displaystyle {\int }_2^6\frac{1}{\rho }d\rho a_z}\left\{\because a_{\rho }\times a_{\varphi }=a_z\right\}\\ =\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }{\left(\ln \rho \right)}_2^6{a}_z\\ =\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }\left(\ln 6-\ln 2\right)a_z\end{array}$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$, 5 for $I_1$, and 2 for $I_2$ in above equation.

$\begin{array}{c}{F}_1=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\right)\left(2\right)}{2\pi }\left(1.0986\right)a_z\\ =2.197\times {10}^{-6}{a}_z\text{N}\\ =2.197a_z\mu \text{N}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the force on side 1 of the triangular loop is $2.197a_z\mu \text{N}$.

(b)

To determine

Find the total force on the loop.

Answer to Problem 13P

The total force on the loop is $0.575a_{\rho }\mu \text{N}$.

Explanation of Solution

Calculation:

Refer to the mentioned Figure given in the textbook.

Write the general expression to calculate the force for side 2.

$F_2={\displaystyle {\oint }_L{I}_{2}d{l}_2\times B_1}$        (2)

Using equation (2), the force on the triangular loop is calculated as follows.

$\begin{array}{c}{F}_2=I_2{\displaystyle \int \left(d\rho a_{\rho }+dz{a}_z\right)\times \left(\frac{{\mu }_{\text{o}}{I}_1}{2\pi \rho }\right)a_{\varphi }}\left\{\because dl=\left(d\rho a_{\rho }+dz{a}_z\right),B_1=\frac{{\mu }_{\text{o}}{I}_1}{2\pi \rho }{a}_{\varphi }\right\}\\ =\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }{\displaystyle \int \frac{1}{\rho }\left(d\rho a_{\rho }+dz{a}_z\right)\times a_{\varphi }}\\ =\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }{\displaystyle \int \frac{1}{\rho }\left(d\rho a_{\rho }\times a_{\varphi }+dz{a}_z\times a_{\varphi }\right)}\end{array}$

$F_2=\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }{\displaystyle \int \frac{1}{\rho }\left(d\rho a_z-dz{a}_{\rho }\right)}\left\{\because a_{\rho }\times a_{\varphi }=a_z,a_z\times a_{\varphi }=-a_{\rho }\right\}$        (3)

Consider,

$\rho =z+2$

On differentiating the above equation,

$d\rho =dz$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$, 5 for $I_1$, and 2 for $I_2$, and $d\rho$ for dz in equation (3) with integration limits.

$\begin{array}{c}{F}_2=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\right)\left(2\right)}{2\pi }{\displaystyle {\int }_4^2\frac{1}{\rho }\left(d\rho a_z-d\rho a_{\rho }\right)}\\ =20\times {10}^{-7}{\displaystyle {\int }_4^2\frac{1}{\rho }d\rho \left(a_z-a_{\rho }\right)}\\ =20\times {10}^{-7}{\left(\ln \rho \right)}_4^2\left(a_z-a_{\rho }\right)\\ =20\times {10}^{-7}\left(\ln 2-\ln 4\right)\left(a_z-a_{\rho }\right)\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{F}_2=20\times {10}^{-7}\left(\ln 2-\ln 4\right)\left(a_z-a_{\rho }\right)\\ =-1.386\times {10}^{-6}\left(a_z-a_{\rho }\right)\text{N}\\ =-1.386\left(a_z-a_{\rho }\right)\mu \text{N}\left\{\because 1\mu ={10}^{-6}\right\}\\ =1.386a_{\rho }-1.386a_z\mu \text{N}\end{array}$

Likewise using equation (3), the force on the triangular loop for side 3 is calculated as follows.

$F_3=\frac{{\mu }_{\text{o}}{I}_1{I}_2}{2\pi }{\displaystyle \int \frac{1}{\rho }\left(d\rho a_z-dz{a}_{\rho }\right)}$        (4)

Consider,

$z=-\rho +6$

On differentiating the above equation,

$dz=-d\rho$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$, 5 for $I_1$, and 2 for $I_2$, and $-d\rho$ for dz in equation (4) with integration limits.

$\begin{array}{c}{F}_3=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\right)\left(2\right)}{2\pi }{\displaystyle {\int }_6^4\frac{1}{\rho }\left(d\rho a_z+d\rho a_{\rho }\right)}\\ =20\times {10}^{-7}{\displaystyle {\int }_6^4\frac{1}{\rho }d\rho \left(a_z+a_{\rho }\right)}\\ =20\times {10}^{-7}{\left(\ln \rho \right)}_6^4\left(a_z+a_{\rho }\right)\\ =20\times {10}^{-7}\left(\ln 4-\ln 6\right)\left(a_z+a_{\rho }\right)\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{F}_3=20\times {10}^{-7}\left(\ln 4-\ln 6\right)\left(a_z+a_{\rho }\right)\\ =-0.8109\times {10}^{-6}\left(a_z+a_{\rho }\right)\\ =-0.8109\times {10}^{-6}{a}_z-0.8109\times {10}^{-6}{a}_{\rho }\\ =-0.8109a_{\rho }-0.8109a_z\mu \text{N}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

The total force on the loop is calculated as follows,

$F=F_1+F_2+F_3$

Substitute $1.386a_{\rho }-1.386a_z\mu \text{N}$ for $F_2$, $-0.8109a_{\rho }-0.8109a_z\mu \text{N}$ for $F_3$, and $2.197a_z\mu \text{N}$ for $F_1$ in above equation.

$\begin{array}{c}F=\left(2.197a_z\mu \text{N}\right)+\left(1.386a_{\rho }-1.386a_z\mu \text{N}\right)+\left(-0.8109a_{\rho }-0.8109a_z\mu \text{N}\right)\\ =0.575a_{\rho }\mu \text{N}\end{array}$

Conclusion:

Thus, the total force on the loop is $0.575a_{\rho }\mu \text{N}$.