Chapter 19, Problem 79P

(a)

To determine

The magnitude and direction of the magnetic field due to wire 1 at the location of wire 2.

Answer to Problem 79P

The magnetic field due to wire 1 at the location of wire 2 is $\underset{\_}{{\overrightarrow{B}}_1=\frac{{\mu }_0{I}_1}{2\pi d}\perp \text{ to the plane of wires}}$.

Explanation of Solution

Given that the separation between wires is $d$, and the length of wires is $L$.

The direction of the magnetic field produced by wire 1 can be determined by the right-hand rule 2. By placing the thumb along the current $I_1$, the curled fingers shows the direction of the magnetic field as shown in Figure 1.

Write the expression for the magnitude of magnetic field at a distance $r$ from a current carrying wire.

$B=\frac{{\mu }_{0}I}{2\pi r}$ (I)

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance.

Conclusion:

From Figure 1, the use of right-hand rule 2 results, the direction of magnetic field due to wire 1 at the location of wire 2 perpendicular to the plane of the wires.

Apply the equation (I) to the given system to obtain the expression for the magnitude of the magnetic field due to wire 1 at the location of wire 2.

$B_1=\frac{{\mu }_0{I}_1}{2\pi d}$

Write the magnetic field due to wire 1 at the location of wire 2 including magnitude and direction.

${\overrightarrow{B}}_1=\frac{{\mu }_0{I}_1}{2\pi d}\perp \text{ to the plane of wires}$

Therefore, the magnetic field due to wire 1 at the location of wire 2 is $\underset{\_}{{\overrightarrow{B}}_1=\frac{{\mu }_0{I}_1}{2\pi d}\perp \text{ to the plane of wires}}$.

(b)

To determine

The magnitude and direction of the magnetic force on wire 2.

Answer to Problem 79P

The magnetic force on wire 2 is $\underset{\_}{\overrightarrow{F}=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}\text{toward }{I}_1}$.

Explanation of Solution

Given that the separation between wires is $d$, and the length of wires is $L$.

The direction of the magnetic force on wire 2 can be determined by the right-hand rule 2 as shown in Figure 1.

Write the expression for the magnetic force due to a current carrying wire.

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

Here, $\overrightarrow{F}$ is the magnetic force, $I$ is the current, $\overrightarrow{L}$ is the length if the wire, and $\overrightarrow{B}$ is the magnetic field.

Conclusion:

From Figure 1, the use of right-hand rule 2 results, the direction of magnetic force on wire 2 towards the current $I_1$.

Apply the equation (II) to the given system to obtain the expression for the magnetic force on wire 2.

$\begin{array}{c}\overrightarrow{F}=I_2\overrightarrow{L}\times {\overrightarrow{B}}_1\\ =I_{2}L{B}_1\text{toward }{I}_1\end{array}$ (III)

Use expression for $B_1$ in (III).

$\begin{array}{c}\overrightarrow{F}=I_{2}L\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\text{toward }{I}_1\\ =\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}\text{toward }{I}_1\end{array}$

Therefore, the magnetic force on wire 2 is $\underset{\_}{\overrightarrow{F}=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}\text{toward }{I}_1}$.

(c)

To determine

The magnitude and direction of the magnetic field due to wire 2 at the location of wire 1.

Answer to Problem 79P

The magnetic field due to wire 2 at the location of wire 1 is $\underset{\_}{{\overrightarrow{B}}_2=\frac{{\mu }_0{I}_2}{2\pi d}\perp \text{ to the plane of wires and opposite to }{\overrightarrow{B}}_1}$.

Explanation of Solution

Given that the separation between wires is $d$, and the length of wires is $L$.

The direction of the magnetic field produced by wire 2 can be determined by the right-hand rule 2. By placing the thumb along the current $I_2$, the curled fingers shows the direction of the magnetic field. The process is similar as done in part (a).

Equation (I) gives expression for the magnitude of magnetic field at a distance $r$ from a current carrying wire.

$B=\frac{{\mu }_{0}I}{2\pi r}$

Conclusion:

The use of right-hand rule 2 results, the direction of magnetic field due to wire 2 at the location of wire 1 perpendicular to the plane of the wires and opposite to ${\overrightarrow{B}}_1$.

Apply the equation (I) to the given system to obtain the expression for the magnitude of the magnetic field due to wire 2 at the location of wire 1.

$B_2=\frac{{\mu }_0{I}_2}{2\pi d}$

Write the magnetic field due to wire 2 at the location of wire 1 including magnitude and direction.

${\overrightarrow{B}}_2=\frac{{\mu }_0{I}_2}{2\pi d}\perp \text{ to the plane of wires and opposite to }{\overrightarrow{B}}_1$

Therefore, the magnetic field due to wire 2 at the location of wire 1 is $\underset{\_}{{\overrightarrow{B}}_2=\frac{{\mu }_0{I}_2}{2\pi d}\perp \text{ to the plane of wires and opposite to }{\overrightarrow{B}}_1}$.

(d)

To determine

The magnitude and direction of the magnetic force on wire 1.

Answer to Problem 79P

The magnetic force on wire 1 is $\underset{\_}{\overrightarrow{F}=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}\text{toward }{I}_2}$.

Explanation of Solution

Given that the separation between wires is $d$, and the length of wires is $L$.

The direction of the magnetic force on wire 1 can be determined by the right-hand rule 2 similar as done in part (b).

Equation (II) gives the expression for the magnetic force due to a current carrying wire.

$\overrightarrow{F}=I\overrightarrow{L}\times \overrightarrow{B}$

Conclusion:

The use of right-hand rule 2 results, the direction of magnetic force on wire 1 towards the current $I_2$.

Apply the equation (II) to the given system to obtain the expression for the magnetic force on wire 1.

$\begin{array}{c}\overrightarrow{F}=I_1\overrightarrow{L}\times {\overrightarrow{B}}_2\\ =I_{1}L{B}_1\text{toward }{I}_2\end{array}$ (III)

Use expression for $B_2$ in (III).

$\begin{array}{c}\overrightarrow{F}=I_{1}L\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\text{toward }{I}_2\\ =\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}\text{toward }{I}_2\end{array}$

Therefore, the magnetic force on wire 1 is $\underset{\_}{\overrightarrow{F}=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}\text{toward }{I}_2}$.

(e)

To determine

Whether the parallel currents in the same direction attract or repel, and whether the parallel currents in opposite direction attract or repel.

Answer to Problem 79P

The parallel currents in the same direction $\underset{\_}{\text{attract}}$, and the parallel currents in opposite direction $\underset{\_}{\text{repel}}$.

Explanation of Solution

The magnetic force is determined from the cross product of the length (along the current direction) and the magnetic field. If the currents in the parallel wires are in the same direction, the forces will be attractive in nature.

Reversing the current’s direction causes the cross products to be oppositely directed. This causes the force on each wire is away from the other wire. Thus, antiparallel currents repels.

Conclusion:

Therefore, the parallel currents in the same direction $\underset{\_}{\text{attract}}$, and the parallel currents in opposite direction $\underset{\_}{\text{repel}}$.

(f)

To determine

Whether the magnitudes and directions of the magnetic force due to the current carrying wires are consistent with Newton’s third law.

Answer to Problem 79P

The magnitudes and directions of the magnetic force due to the current carrying wires are consistent with Newton’s third law.

Explanation of Solution

The forces on the two currents are equal in magnitude and opposite in direction. According to Newton’s third law, for every action there is an equal and opposite reaction. In the case of current passing through the parallel wires, the forces can be identified as one opposes the other but equal in magnitude.

Both the parallel currents and antiparallel currents are consistent with Newton’s third law, since the forces on the two currents are equal and opposite in direction.

Conclusion:

Therefore, the magnitudes and directions of the magnetic force due to the current carrying wires are consistent with Newton’s third law.