Chapter 19, Problem 87P

(a)

To determine

The separation between the wires at equilibrium.

Answer to Problem 87P

The separation between the wires at equilibrium is $4.9\text{cm}$.

Explanation of Solution

Refer figure 1.

Write an expression for the forces acting on a wire.

    $\begin{array}{c}\Sigma F_x=2T\sin \theta -F\\ =0\end{array}$                                                                                                (I)

Here, $\Sigma F_x$ is the x component of the net force on one wire at equilibrium, $T$ is the tension on the string, $\theta$ is the angle and $F$ is the force on the string.

Rewrite the expression (I) to find $T$.

    $T=\frac{F}{2\sin \theta }$                                                                                                         (II)

Write an expression for the Newton’s second law for one wire.

    $\begin{array}{c}\Sigma F_y=2T\cos \theta -mg\\ =0\end{array}$                                                                                          (III)

Here, $\Sigma F_y$ is y component of the net force on one wire at equilibrium, $m$ is the mass and $g$ is the acceleration due to gravity.

Rewrite the expression (III) to find $T$.

    $T=\frac{mg}{2\cos \theta }$                                                                                                       (IV)

The forces are repelling. Thus, the current in the wires will be opposite in direction.

Equate equation (II) and (IV).

    $\frac{F}{2\sin \theta }=\frac{mg}{2\cos \theta }$                                                                                                (V)

Rearrange the equation (V) to find $F$.

    $F=mg\tan \theta$                                                                                                   (VI)

Refer figure 1 to rewrite equation (VI).

    $F=mg\tan \theta$

For small angle approximation $\tan \theta \approx \sin \theta$. The opposite side of the angle $\theta$ is half the distance between the wires $r$ and the adjacent side to the angle $\theta$ is the length of the strings $L$.

Substitute $\frac{1}{2}r$ for the opposite side and $L$ for the adjacent and use the small angle approximation to find $F$.

    $\begin{array}{c}\tan \theta =\frac{\text{opposite side}}{\text{adjacent side}}\\ =\frac{\frac{1}{2}r}{L}\\ \sin \theta =\frac{\frac{1}{2}r}{L}\\ =\frac{r}{2L}\end{array}$

  $\begin{array}{c}F=mg\sin \theta \\ =mg\left(\frac{\frac{1}{2}r}{L}\right)\\ =mg\left(\frac{r}{2L}\right)\end{array}$                                                                                           (VII)

Here, $L$ is the length of the wire and $r$ is the distance between the wires at equilibrium.

Write an expression for the force due to the wires.

    $F=\frac{{\mu }_{0}d{I}_1{I}_2}{2\pi r}$                                                                                                   (VIII)

Here, ${\mu }_0$ is the permeability of free space, $d$ is the separation between the wires, $I_1$ is the current in first wire and $I_2$ is the current in second wire.

Equate equation (VII) and (VIII).

  $\frac{{\mu }_{0}d{I}_1{I}_2}{2\pi r}=mg\left(\frac{r}{2L}\right)$                                                                                           (IX)

Rearrange the equation to find $r$.

    $r=\sqrt{\frac{{\mu }_0{I}_1{I}_{2}L}{\left(\frac{m}{d}\right)g\pi }}$

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $25.0\text{A}$ for $I_1$, $100.0\text{A}$ for $I_2$, $1.2\text{m}$ for $L$, $9.80{\text{m/s}}^2$ for $g$ and $0.050\text{kg/m}$ for $m/d$ in equation (IX) to find $r$.

    $\begin{array}{c}r=\sqrt{\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(25.0\text{A}\right)\left(100.0\text{A}\right)\left(1.2\text{m}\right)}{\left(0.050\text{kg/m}\right)\left(9.80{\text{m/s}}^2\right)\pi }}\\ =\left(4.9\times {10}^{-2}\text{m}\right)\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)\\ =4.9\text{cm}\end{array}$

Thus, the separation between the wires at equilibrium is $4.9\text{cm}$.

(b)

To determine

The direction of the current in the wires.

Answer to Problem 87P

The currents in the wires will be opposite in directions.

Explanation of Solution

Write an expression for the force due to the current in the wire.

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

Here, $\overrightarrow{F}$ is the force, $I$ is the current through the wire, $\overrightarrow{d}$ is the length of the strings and $\overrightarrow{B}$ is the magnetic field.

According to Fleming’s right hand rule, if the thumb, the index finger and the middle finger of the right hand are kept mutually perpendicular to each other and if the middle finger points in the direction of magnetic field, and the index finger points in the direction of current, then the thumb points in the direction of the force. Thus, to get a repulsive force, the current through the wires should be opposite to each other.

(c)

To determine

If the forces in the wires are consistent with Newton’s third law of motion.

Answer to Problem 87P

Yes, the forces in the wires are consistent with Newton’s third law of motion.

Explanation of Solution

The currents in the wires are different. The magnitude of the magnetic fields differs with the magnitude of current through the wires. Thus, the fields created by current through the wires are different.

However, the forces acting on the wires are the same in magnitude and opposite in direction. The wires hang away from the vertical at equal angles. Thus, the forces in the wires are consistent with Newton’s third law of motion.