Chapter 26, Problem 52P

(a)

To determine

Whether the magnetic moment of the coil make angle with the unit vector $\widehat{i}$ .

Answer to Problem 52P

The coil makes an angle of $\widehat{i}$ with the y axis.

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The angel that the coil makes with the y axis is $\widehat{n}$ and the normal is drawn to the coil along the x axis.

In the Figure 1, the angel that the coil makes with y axis is $\widehat{i}$ . The normal drawn to the coil have an angle $\widehat{j}$ below the positive x-axis.

Conclusion:

Therefore, the coil makes an angle of $\widehat{i}$ with the y axis.

(b)

To determine

The expression for $\widehat{n}$ in terms of other unit vectors.

Answer to Problem 52P

The expression for $\widehat{n}$ is $0.8\widehat{i}-0.6\widehat{j}$ .

Explanation of Solution

Given:

The angle $\theta$ is $37°$ .

Formula:

The normal drawn to the coil in the plane of positive $x$ and negative $y$ direction and the expression for $\widehat{n}$ is given by,

  $\begin{array}{c}\widehat{n}=n_x\widehat{i}-n_y\widehat{j}\\ =\cos \theta \widehat{i}-\sin \theta \widehat{j}\end{array}$

Calculation:

The value of $\widehat{n}$ is calculated as,

  $\begin{array}{c}\widehat{n}=\cos \left(37°\right)\widehat{i}-\sin \left(37°\right)\widehat{j}\\ =0.799\widehat{i}-0.692\widehat{j}\\ =0.8\widehat{i}-0.6\widehat{j}\end{array}$

Conclusion:

Therefore, the expression for $\widehat{n}$ is $0.8\widehat{i}-0.6\widehat{j}$ .

(c)

To determine

The magnetic moment of the coil.

Answer to Problem 52P

The value of the magnetic field is $\left(0.4185\text{N}\cdot \text{m}\right)\widehat{k}$ .

Explanation of Solution

Given:

The length of the coil $l$ is $5.00\text{cm}$

The width of the coil $w$ is $8\text{cm}$ .

The current $I$ in the loop is $1.75\text{A}$ .

The magnetic field density $\overrightarrow{B}$ is $1.5\text{T}$ .

Formula:

The expression for the area of the loop is given by,

  $A=lw$

The expression to determine the value of $\overrightarrow{\mu }$ is given by,

  $\overrightarrow{\mu }=INA\widehat{n}$

The expression to determine the magnetic moment of the coil is given by,

  $\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$

Calculation:

The area of the loop is calculated as,

  $\begin{array}{c}A=lw\\ =\left(5\text{cm}\right)\left(8\text{cm}\right)\\ =40{\text{cm}}^2\end{array}$

The value of $\overrightarrow{\mu }$ is calculated as,

  $\begin{array}{c}\overrightarrow{\mu }=INA\widehat{n}\\ =\left(1.75\text{A}\right)\left(40{\text{cm}}^2\right)\left(50\right)\left(0.799\widehat{i}-0.692\widehat{j}\right)\\ =\left(0.279\text{A}\cdot {\text{m}}^2\right)\widehat{i}-\left(0.21\text{A}\cdot {\text{m}}^2\right)\widehat{j}\end{array}$

The magnetic moment of the coil is calculated as,

  $\begin{array}{c}\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}\\ =\left[\left(0.279\text{A}\cdot {\text{m}}^2\right)\widehat{i}-\left(0.21\text{A}\cdot {\text{m}}^2\right)\widehat{j}\right]\times \left[1.5\text{T}\right]\widehat{j}\\ =\left(0.4185\text{N}\cdot \text{m}\right)\widehat{k}\end{array}$

Conclusion:

Therefore, the value of the magnetic moment is $\left(0.4185\text{N}\cdot \text{m}\right)\widehat{k}$ .

(d)

To determine

The torque on the coil when the magnetic field is constant.

Answer to Problem 52P

The torque on the coil is $0.315\text{J}$ .

Explanation of Solution

Formula:

The expression for the torque on the coil is given by,

  $U=-\overrightarrow{\mu }\overrightarrow{B}$

Calculation:

The value of the torque on the coil is calculated as,

  $\begin{array}{c}U=-\overrightarrow{\mu }\overrightarrow{B}\\ =-\left[\left(0.279\text{A}\cdot {\text{m}}^2\right)\widehat{i}-\left(0.21\text{A}\cdot {\text{m}}^2\right)\widehat{j}\right]\left(1.5\text{T}\right)\widehat{j}\\ =0.315\text{J}\end{array}$

Conclusion:

Therefore, the torque on the coil is $0.315\text{J}$ .

(e)

To determine

The potential energy of the coil.

Answer to Problem 52P

The potential energy of the coil is $0.315\text{J}$ .

Explanation of Solution

Formula:

The expression to determine the potential energy of the coil is given by,

  $U=-\overrightarrow{\mu }\overrightarrow{B}$

Calculation:

The potential energy of the coil is calculated as,

  $\begin{array}{c}U=-\overrightarrow{\mu }\overrightarrow{B}\\ =-\left[\left(0.279\text{A}\cdot {\text{m}}^2\right)\widehat{i}-\left(0.21\text{A}\cdot {\text{m}}^2\right)\widehat{j}\right]\left(1.5\text{T}\right)\widehat{j}\\ =0.315\text{J}\end{array}$

Conclusion:

Therefore, the potential energy of the coil is $0.315\text{J}$ .