Chapter 19, Problem 50P

(a)

To determine

What is the magnetic force on each side of the loop if the magnetic field is $2.5\text{T}$ out of the page?

Answer to Problem 50P

Magnetic force on the top side of loop is $0.75\text{N}$ in negative y direction, force on bottom side of loop is $0.75\text{N}$ in the positive y direction, force on left side of loop is $0.50\text{N}$ in the positive x direction, and force on the right side of the loop is $0.50\text{N}$ in the negative x direction.

Explanation of Solution

The schematic diagram showing the rectangular loop carrying current is given in figure 1 below. The direction of magnetic field is given to be out of the page. The four sides of the loop is also marked in the figure. Consider each side separately to find the magnitude and direction of force experienced in the loop.

Write the equation to find the force on the top side of the loop.

${\overrightarrow{F}}_{top}=I{\overrightarrow{L}}_{top}\times \overrightarrow{B}$ (I)

Here, ${\overrightarrow{F}}_{top}$ is the force on top side of loop, $I$ is the current in the loop, ${\overrightarrow{L}}_{top}$ is the length of the top side, $\overrightarrow{B}$ is the magnetic field vector

Write the equation to find the force on the bottom side of the loop.

${\overrightarrow{F}}_{bottom}=I{\overrightarrow{L}}_{bottom}\times \overrightarrow{B}$ (II)

Here, ${\overrightarrow{F}}_{top}$ is the force on bottom side of loop, $I$ is the current in the loop, ${\overrightarrow{L}}_{top}$ is the length of the bottom side, $\overrightarrow{B}$ is the magnetic field vector

Write the equation to find the force on the left side of the loop.

${\overrightarrow{F}}_{left}=I{\overrightarrow{L}}_{left}\times \overrightarrow{B}$ (III)

Here, ${\overrightarrow{F}}_{left}$ is the force on left side of loop, $I$ is the current in the loop, ${\overrightarrow{L}}_{left}$ is the length of the left side, $\overrightarrow{B}$ is the magnetic field vector

Write the equation to find the force on the right side of the loop.

${\overrightarrow{F}}_{right}=I{\overrightarrow{L}}_{right}\times \overrightarrow{B}$ (IV)

Here, ${\overrightarrow{F}}_{right}$ is the force on right side of loop, $I$ is the current in the loop, ${\overrightarrow{L}}_{right}$ is the length of the right side, $\overrightarrow{B}$ is the magnetic field vector

Conclusion

Substitute $1.0\text{A}$ for $I$ , $0.300\text{m right}$ for $L_{top}$ , $2.5\text{T}\text{out of page}$ for $B$ in equation (I) to get $F_{top}$

$\begin{array}{c}{F}_{top}=\left(1.0\text{A}\right)\left(0.300\text{m}\text{right}\right)\times \left(2.5\text{T}\text{out of page}\right)\\ =0.75\text{N in the -y direction}\end{array}$

Substitute $1.0\text{A}$ for $I$ , $0.300\text{m left}$ for $L_{top}$ , $2.5\text{T}\text{out of page}$ for $B$ in equation (II) to get $F_{bottom}$

$\begin{array}{c}{F}_{bottom}=\left(1.0\text{A}\right)\left(0.300\text{m}left\right)\times \left(2.5\text{T}\text{out of page}\right)\\ =0.75\text{N in the +y direction}\end{array}$

Substitute $1.0\text{A}$ for $I$ , $0.300\text{m up}$ for $L_{top}$ , $2.5\text{T}\text{out of page}$ for $B$ in equation (III) to get $F_{left}$

$\begin{array}{c}{F}_{left}=\left(1.0\text{A}\right)\left(0.300\text{m}up\right)\times \left(2.5\text{T}\text{out of page}\right)\\ =0.50\text{N in the +x direction}\end{array}$

Substitute $1.0\text{A}$ for $I$ , $0.300\text{m down}$ for $L_{right}$ , $2.5\text{T}\text{out of page}$ for $B$ in equation (IV) to get $F_{right}$

$\begin{array}{c}{F}_{right}=\left(1.0\text{A}\right)\left(0.300\text{m}down\right)\times \left(2.5\text{T}\text{out of page}\right)\\ =0.50\text{N in the -x direction}\end{array}$

Therefore, Magnetic force on the top side of loop is $0.75\text{N}$ in negative y direction, force on bottom side of loop is $0.75\text{N}$ in the positive y direction, force on left side of loop is $0.50\text{N}$ in the positive x direction, and force on the right side of the loop is $0.50\text{N}$ in the negative x direction.

(b)

To determine

What is the net magnetic force on the loop?

Answer to Problem 50P

The net magnetic force on the loop is $0\text{N}$.

Explanation of Solution

The net force is the resultant sum of forces in the x and y directions.

Write the equation to find the net force in x direction.

$F_{net,x}=F_{left}+F_{right}$ (I)

Here, $F_{net,x}$ is the net force in the x direction, $F_{left}$ is the force in the left side of loop, $F_{right}$ is the force in the right side of the loop

Write the equation to find the net force in y direction.

$F_{net,y}=F_{top}+F_{bottom}$ (II)

Here, $F_{net,y}$ is the net force in the y direction, $F_{top}$ is the force in the top side of loop, $F_{bottom}$ is the force in the bottom side of the loop

Write the equation to find the sum of x and y component of forces.

$F_{net}=F_{net,x}+F_{net,y}$ (III)

Here, $F_{net}$ is the net force in the loop, $F_{net,x}$ is the net force in the x direction, $F_{net,y}$ is the net force in the y direction

Conclusion:

Substitute $0.50\text{N}$ for $F_{left}$ , and $-0.50\text{N}$ for $F_{right}$ in equation (I) to get $F_{net,x}$

$\begin{array}{c}{F}_{net,x}=0.50\text{N-0}\text{.50}\text{N}\\ \text{=0}\text{N}\end{array}$

Substitute $0.75\text{N}$ for $F_{bottom}$ , and $-0.75\text{N}$ for $F_{top}$ in equation (II) to get $F_{net,y}$

$\begin{array}{c}{F}_{net,y}=0.75\text{N-0}\text{.75}\text{N}\\ \text{=0}\text{N}\end{array}$

Substitute $0\text{N}$ for $F_{net,x}$ and $F_{net,y}$ in equation (III) to get $F_{net}$

$\begin{array}{c}{F}_{net}=0\text{N+0}\text{N}\\ \text{=0}\text{N}\end{array}$

Therefore, The net magnetic force on the loop is $0\text{N}$.