Chapter 19, Problem 109P

(a)

To determine

The direction of the magnetic force on each of the four sides of the rectangle due to the long wire’s magnetic field.

Answer to Problem 109P

The direction of magnetic force each side of the rectangle is given in below table.

Side	Current direction	Field direction	Force direction
Top	right	Out of the page	Down: attracted to long wire
Bottom	Left	Out of the page	Up: repelled by long wire
Left	Up	Out of the page	Right
Right	down	Out of the page	left

Explanation of Solution

Write the expression for the magnitude of magnetic field due to current carrying long wire.

$B=\frac{{\mu }_0{I}_2}{2\pi r}$

Here, $B$ is the magnitude of magnetic field at a radial distance $r$ from the current carrying long wire, $I_2$ is the current flowing through wire and $r$ is radial distance of the point where the magnetic field is to be calculated.

The direction of magnetic field is given by right hand rule. According to the rule, when the thumb is pointed in the direction of current and the fingers are curled, the direction of fingers represents the direction of magnetic field lines. The tangent of the field line at any point gives the direction of magnetic field at that point.

Since current flows to right, magnetic field points out of the page.

Write the expression for force acting on one side of current carrying rectangular loop of wire.

$\overrightarrow{F}=I_1\overrightarrow{L}\times \overrightarrow{B}$ (I)

Here, $\overrightarrow{F}$ is the magnetic force acting on one side of current carrying rectangular loop of wire, $I_1$ is the current through the wire, $\overrightarrow{L}$ is the length of the wire at which force is acting with direction same as direction of current through wire and $\overrightarrow{B}$ is the magnetic field due to current carrying long wire.

The direction of force is given by the direction of $\overrightarrow{L}\times \overrightarrow{B}$.

Consider the bottom side of rectangular loop, where current is flowing to the left and magnetic field is out of the page. According to right hand rule, magnetic force acts in the upward direction. That is repelled by long wire.

Consider the top side of rectangular loop, where current is flowing to the right and magnetic field is out of the page. According to right hand rule, magnetic force acts in the downward direction. That is attracted to the long wire.

Consider the left side of rectangular loop, where current is flowing up and magnetic field is out of the page. According to right hand rule, magnetic force acts towards the right.

Consider the right side of rectangular loop, where current is flowing downward and magnetic field is out of the page. According to right hand rule, magnetic force acts toward the left.

Conclusion:

Therefore, the direction of magnetic force each side of the rectangle is given in below table.

Side	Current direction	Field direction	Force direction
Top	right	Out of the page	Down: attracted to long wire
Bottom	Left	Out of the page	Up: repelled by long wire
Left	Up	Out of the page	Right
Right	down	Out of the page	left

(b)

To determine

The net magnetic force on the rectangular loop due to the long wire’s magnetic field.

Answer to Problem 109P

The net magnetic force on the rectangular loop due to the long wire’s magnetic field is $1.0\times {10}^{-8}\text{N}$ away from the wire.

Explanation of Solution

The magnetic field along the left and right side of the rectangular loop have same magnitude at each point of wire. The left and right side of the rectangular loop experience equal magnitude of magnetic force, since the two sides are symmetrically situated with respected to long wire. The top of the loop experiences small magnetic field than bottom side, since radial distance of top side is larger than that of the bottom side.

Since magnetic forces on left and right side of the loop  are equal in magnitude and opposite in direction, they cancel each other.

Write the expression for the net force acting on the rectangular loop.

$F=F_{\text{bottom}}-F_{\text{top}}$ (II)

Here, $F$ is the magnitude of net force acting on the wire, $F_{\text{bottom }}$ is the magnitude of magnetic force on bottom side of the loop and $F_{\text{top}}$ is the magnitude of magnetic force on top side of the loop.

The negative sign indicates that force on top and bottom side are opposite in direction.

Write the expression to calculate magnitude of magnetic force on each side of wire.

$F_{\text{side}}=I_{1}LB$

Here, $F_{\text{side}}$ is the magnitude of magnetic force on each side, $L$ is the length of wire segment and $B$ is the magnetic field.

From equation (I), write the expression for the magnitude of magnetic force on bottom wire.

$F_{\text{bottom}}=I_1{L}_{\text{bottom}}{B}_{\text{bottom}}$ (III)

Here, $L_{\text{bottom }}$ is the length of bottom side of rectangular loop and $B_{\text{bottom }}$ is the magnitude of magnetic field acting on the bottom side of rectangular loop.

From equation (I), write the expression for the magnitude of magnetic force on top wire.

$F_{\text{top}}=I_1{L}_{\text{top}}{B}_{\text{top}}$ (IV)

Here, $L_{\text{top}}$ is the length of top side of rectangular loop, $B_{\text{top}}$ is the magnitude of magnetic field acting on the top side of rectangular loop.

Write the expression for $B_{\text{bottom}}$.

$B_{\text{bottom}}=\frac{{\mu }_0{I}_2}{2\pi r_{\text{bottom}}}$

Here, ${\mu }_0$ is the permeability of free space and $r_{\text{bottom}}$ is the radial distance between the bottom of the rectangular loop and the long wire.

Write the expression for $B_{\text{top}}$.

$B_{\text{top}}=\frac{{\mu }_0{I}_2}{2\pi r_{\text{top}}}$

Here, $r_{\text{top}}$ is the radial distance between the top side of the rectangular loop and the long wire.

Substitute $\frac{{\mu }_0{I}_2}{2\pi r_{\text{bottom}}}$ for $B_{\text{bottom}}$ and $L$ for $L_{\text{botttom}}$ in equation (III) to get $F_{\text{bottom}}$.

$F_{\text{bottom}}=I_{1}L\frac{{\mu }_0{I}_2}{2\pi r_{\text{bottom}}}$

Here, $L$ is the length of top and bottom side of rectangular loop.

Substitute $\frac{{\mu }_0{I}_2}{2\pi r_{\text{top}}}$ for $B_{\text{top}}$ and $L$ for $L_{\text{botttom}}$ in equation (IV) to get $F_{\text{top}}$.

$F_{\text{top}}=I_{1}L\frac{{\mu }_0{I}_2}{2\pi r_{\text{top}}}$

Substitute $I_{1}L\frac{{\mu }_0{I}_2}{2\pi r_{\text{bottom}}}$ for $F_{\text{bottom}}$ and $I_{1}L\frac{{\mu }_0{I}_2}{2\pi r_{\text{top}}}$ for $F_{\text{top}}$ in equation (II) to get net force on loop.

$\begin{array}{l}F=I_{1}L\frac{{\mu }_0{I}_2}{2\pi r_{\text{bottom}}}-I_{1}L\frac{{\mu }_0{I}_2}{2\pi r_{\text{top}}}\\ =\frac{I_1{I}_{2}L{\mu }_0}{2\pi }\left(\frac{1}{r_{\text{bottom}}}-\frac{1}{r_{\text{top}}}\right)\end{array}$ (V)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $2.0\text{mA}$ for $I_1$ , $8.0\text{A}$ for $I_2$ , $9.0\text{ cm}$ for $L$ , $2.0\text{cm}$ for $r_{\text{bottom}}$ and $7.0\text{cm}$ for $r_{\text{top}}$ in equation (V) to get $F$.

$\begin{array}{c}F=\frac{\left(2.0\text{mA}\times \frac{1\text{A}}{1000\text{mA}}\right)\left(8.0\text{A}\right)\left(9.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)}{2\pi }\left(\frac{1}{2.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}-\frac{1}{7.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)\\ =1.0\times {10}^{-8}\text{N}\times \frac{{10}^9\text{nN}}{1\text{N}}\\ =10\text{nN}\end{array}$

Since answer is positive, the net force must direct along the direction of force acting on the top side of rectangular loop. Therefore, net force is directed away from the long wire.

Therefore, the net magnetic force on the rectangular loop due to the long wire’s magnetic field is $1.0\times {10}^{-8}\text{N}$ away from the wire.

(c)

To determine

The magnetic force on the long wire due to the loop.

Answer to Problem 109P

The magnetic force on the long wire due to the loop is $10\text{nN}$ down in the diagram, away from the loop.

Explanation of Solution

According to Newton’s third law force experienced by current carrying rectangular wire due to long wire is equal and opposite to force experienced by long wire due to rectangular wire.

The net magnetic force on the rectangular loop due to the long wire’s magnetic field is $10\text{nN}$ away from the wire. Therefore, force acting on the current carrying long wire is equal to $10\text{nN}$ and is directed in downward direction.

Conclusion:

The net magnetic force on the rectangular loop due to the long wire’s magnetic field is $10\text{nN}$ away from the wire. Therefore, force acting on the current carrying long wire is equal to $10\text{nN}$ and is directed in downward direction.