Chapter 8, Problem 8.18P

To determine

(a)

The vector torque on the given square loop.

Answer to Problem 8.18P

The vector torque on the square loop is, $T=-0.96a_x\text{N}\cdot \text{m}$.

Explanation of Solution

Given Information:

Square loop is given as,.

.

The torque is about the origin $A\left(0,0,0\right)$ in the field $B=100a_y\text{mT}$ ..

Calculation:

The area of loop,

   $\begin{array}{c}S=\left(2-\left(-2\right)\right)a_x\times \left(2-\left(-2\right)\right)a_y\\ =16a_z{\text{m}}^2\end{array}$.

The field given is uniform. So, the torque about any origin is independent to that origin point.

Thus, the torque about $A\left(0,0,0\right)$.

   $\begin{array}{c}T=\left(IS\right)\times B\\ =0.6\left(16a_z\right)\times \left(100a_y\right){10}^{-3}\\ =-0.96a_x\text{N}\cdot \text{m}\end{array}$.

Conclusion:

The vector torque on the square loop is, $T=-0.96a_x\text{N}\cdot \text{m}$.

To determine

(b)

The vector torque on the given square loop.

Answer to Problem 8.18P

The vector torque on the square loop is, $T=-0.96a_x+1.92a_y\text{N}\cdot \text{m}$.

Explanation of Solution

Given Information:

Square loop is given as,.

.

The torque is about the origin $A\left(0,0,0\right)$ in the field $B=200a_x+100a_y\text{mT}$ ..

Calculation:

The area of loop,

   $\begin{array}{c}S=\left(2-\left(-2\right)\right)a_x\times \left(2-\left(-2\right)\right)a_y\\ =16a_z{\text{m}}^2\end{array}$.

The field given is uniform. So, the torque about any origin is independent to that origin point.

Thus the torque about $A\left(0,0,0\right)$ ,

   $\begin{array}{c}T=\left(IS\right)\times B\\ =0.6\left(16a_z\right)\times \left(200a_x+100a_y\right){10}^{-3}\\ =-0.96a_x+1.92a_y\text{N}\cdot \text{m}\end{array}$.

Conclusion:

The vector torque on the square loop is, $T=-0.96a_x+1.92a_y\text{N}\cdot \text{m}$.

To determine

(c)

The vector torque on the given square loop.

Answer to Problem 8.18P

The vector torque on the square loop is, $T=-0.96a_x+1.92a_y\text{N}\cdot \text{m}$.

Explanation of Solution

Given Information:

Square loop is given as,.

.

The torque is about the origin $A\left(1,2,3\right)$ in the field $B=200a_x+100a_y-300a_z\text{mT}$ ..

Calculation:

The area of loop,

   $\begin{array}{c}\text{S}=\left(2-\left(-2\right)\right)a_x\times \left(2-\left(-2\right)\right)a_y\\ =16a_z{\text{m}}^2\end{array}$.

The field given is uniform. So, the torque about any origin is independent to that origin point.

Thus the torque about $A\left(1,2,3\right)$ ,

   $\begin{array}{c}\text{T}=\left(IS\right)\times B\\ =0.6\left(16a_z\right)\times \left(200a_x+100a_y-300a_z\right)\times {10}^{-3}\\ =-0.96a_x+1.92a_y\text{N}\cdot \text{m}\end{array}$.

Conclusion:

The vector torque on the square loop is, $T=-0.96a_x+1.92a_y\text{N}\cdot \text{m}$.

To determine

(d)

The vector torque on the given square loop.

Answer to Problem 8.18P

The vector torque on the square loop is, $T=0.96a_x+2.64a_y-1.44a_z\text{N}\cdot \text{m}$.

Explanation of Solution

Given Information:

Square loop is given as,.

The torque is about the origin $A\left(1,2,3\right)$ in the field $B=200a_x+100a_y-300a_z\text{mT}$ for $x\ge 2$ and $B=0$ elsewhere..

Calculation:

Since the magnetic field is present only for $x\ge 2$ , the force is acting only on theydirected segment at $x=2$ ..

The differential force,

   $\begin{array}{c}d\text{F}=\left(Id\text{L}\right)\times \text{B}\\ =0.6\left(dy{\text{a}}_y\right)\times \left(200a_x+100a_y-300a_z\right)\times {10}^{-3}\\ =\left(-0.18a_x-0.12a_z\right)dy\end{array}$.

The position vector of any point in the segment, relative to given origin,

   $\begin{array}{c}\text{R}=\left(2,y,0\right)-\left(1,2,3\right)\\ =a_x+\left(y-2\right)a_y-3a_z\end{array}$.

So, the differential torque,

   $\begin{array}{c}d\text{T}=\text{R}\times d\text{F}\\ =\left(a_x+\left(y-2\right)a_y-3a_z\right)\times \left(-0.18a_x-0.12a_z\right)dy\\ =\left(-0.12\left(y-2\right)a_x+0.66a_y+0.18\left(y-2\right)a_z\right)dy\end{array}$.

The total torque,

   $\begin{array}{c}T={\displaystyle \int d\text{T}}\\ ={\displaystyle \underset{-2}{\overset{2}{\int }}\left(-0.12\left(y-2\right)a_x+0.66a_y+0.18\left(y-2\right)a_z\right)dy}\\ =-0.12{\left[\frac{y^2}{2}-2y\right]}_{-2}^2{a}_x+0.66{\left[y\right]}_{-2}^2{a}_y+0.18{\left[\frac{y^2}{2}-2y\right]}_{-2}^2{a}_z\\ =0.96a_x+2.64a_y-1.44a_z\text{N}\cdot \text{m}\end{array}$.

Conclusion:

The vector torque on the square loop is, $T=0.96a_x+2.64a_y-1.44a_z\text{N}\cdot \text{m}$.