Chapter 19, Problem 51P

(a)

To determine

The magnetic field on each side of the loop.

Answer to Problem 51P

The magnetic force on each side of the loop is $\underset{\_}{{\overrightarrow{F}}_{\text{top}}=0}$, $\underset{\_}{{\overrightarrow{F}}_{\text{bottom}}=0}$, $\underset{\_}{{\overrightarrow{\text{F}}}_{\text{left}}\text{=0}\text{.50}\text{N out o the page}}$, $\underset{\_}{{\overrightarrow{F}}_{\text{right}}=0.50\text{N}\text{ into the page}}$.

Explanation of Solution

The rectangle wire loop is given below.

Write the expression for magnetic force along top loop.

$\begin{array}{c}{\overrightarrow{F}}_{\text{top}}=I{\overrightarrow{L}}_{\text{top}}\times \overrightarrow{B}\\ =I{\overrightarrow{L}}_{\text{top}}\overrightarrow{B}\sin \theta \end{array}$ (I)

Here, ${\overrightarrow{F}}_{\text{top}}$ is the magnetic force along top side, $I$ is the current, $B$ is the magnetic field, ${\overrightarrow{L}}_{\text{top}}$ is the length of the top of the wire, and $\theta$ is the angle between force and field.

Write the expression for magnetic force along the bottom of the wire.

$\begin{array}{c}{\overrightarrow{F}}_{\text{bottom}}=I{\overrightarrow{L}}_{\text{bottom}}\times \overrightarrow{B}\\ =-I{\overrightarrow{L}}_{\text{top}}\overrightarrow{B}\sin \theta \end{array}$ (II)

Here, ${\overrightarrow{L}}_{\text{bottom}}$ is the length of the wire on bottom side.

The negative sign indicates the direction of magnetic field on top and bottom of the wire is opposite to each other.

Write the expression for magnetic force along the left of the wire.

$\begin{array}{c}{\overrightarrow{F}}_{\text{left}}=I{\overrightarrow{L}}_{\text{left}}\times \overrightarrow{B}\\ =I{\overrightarrow{L}}_{\text{left}}\overrightarrow{B}\sin \theta \end{array}$ (III)

Here, ${\overrightarrow{L}}_{\text{left}}$ is the length of the wire on left side.

Similar to equation (II) the expression for force along right side of the wire.

$\begin{array}{c}{\overrightarrow{F}}_{\text{right}}=I{\overrightarrow{L}}_{\text{right}}\times \overrightarrow{B}\\ =-I{\overrightarrow{L}}_{\text{left}}\overrightarrow{B}\sin \theta \end{array}$ (IV)

Conclusion:

Substitute, $1.0\text{A}$ for $I$, $0.300\text{m}$ for ${\overrightarrow{L}}_{\text{top}}$, and $2.5\text{T}$ for $B$, and $180°$ for $\theta$ in equation (I).

$\begin{array}{c}{\overrightarrow{F}}_{\text{top}}=1.0\text{A}\times \text{0}\text{.300}\text{m}\times \text{2}\text{.5}\text{T}\times \sin \left(180°\right)\\ =0\end{array}$

Substitute, $1.0\text{A}$ for $I$, $0.300\text{m}$ for ${\overrightarrow{L}}_{\text{top}}$, and $2.5\text{T}$ for $B$, and $0°$ for $\theta$ in equation (II).

$\begin{array}{c}{\overrightarrow{F}}_{\text{top}}=1.0\text{A}\times \text{0}\text{.300}\text{m}\times \text{2}\text{.5}\text{T}\times \sin \left(0°\right)\\ =0\end{array}$

Substitute, $1.0\text{A}$ for $I$, $0.200\text{m}$ for ${\overrightarrow{L}}_{\text{top}}$, and $2.5\text{T}$ for $B$, and $90°$ for $\theta$ in equation (III).

$\begin{array}{c}{\overrightarrow{F}}_{\text{top}}=1.0\text{A}\times \text{0}\text{.200}\text{m}\times \text{2}\text{.5}\text{T}\times \sin \left(90°\right)\\ =0.5\text{N}\end{array}$

Substitute, $1.0\text{A}$ for $I$, $0.200\text{m}$ for ${\overrightarrow{L}}_{\text{top}}$, and $2.5\text{T}$ for $B$, and $90°$ for $\theta$ in equation (IV).

$\begin{array}{c}{\overrightarrow{F}}_{\text{top}}=1.0\text{A}\times \text{0}\text{.200}\text{m}\times \text{2}\text{.5}\text{T}\times \sin \left(90°\right)\\ =0.5\text{N}\end{array}$

The force along left side of the wire is directed out of the page, and force along right side of the wire is directed into the page.

Therefore, the magnetic force on each side of the loop is $\underset{\_}{{\overrightarrow{F}}_{\text{top}}=0}$, $\underset{\_}{{\overrightarrow{F}}_{\text{bottom}}=0}$, $\underset{\_}{{\overrightarrow{\text{F}}}_{\text{left}}\text{=0}\text{.50}\text{N out o the page}}$, $\underset{\_}{{\overrightarrow{F}}_{\text{right}}=0.50\text{N}\text{ into the page}}$.

(b)

To determine

The net magnetic force on the loop.

Answer to Problem 51P

The net magnetic force on the loop is $\underset{\_}{0}$.

Explanation of Solution

The net force is the sum of the components of forces.

Write the expression for net magnetic force.

${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_{\text{top}}+{\overrightarrow{F}}_{\text{bottom}}+{\overrightarrow{F}}_{\text{left}}+{\overrightarrow{F}}_{\text{right}}$ (V)

Conclusion:

Substitute, $0$ for ${\overrightarrow{F}}_{\text{top}}$, $0$ for ${\overrightarrow{F}}_{\text{bottom}}$, $0.50\text{N }$ for ${\overrightarrow{F}}_{\text{left}}$, and $-0.50\text{N}$ for ${\overrightarrow{F}}_{\text{right}}$ in equation (V).

$\begin{array}{c}{\overrightarrow{F}}_{\text{net}}=0+0+0.50\text{N}+\left(-050\text{N}\right)\\ =0\end{array}$

Therefore, net magnetic force on the loop is $\underset{\_}{0}$.