Chapter 30, Problem 91PQ

Three long, current-carrying wires are parallel to one another and separated by a distance d. The magnitudes and directions of the currents are shown in Figure P30.91. Wires 1 and 3 are fixed, but wire 2 is free to move. Wire 2 is displaced to the right by a small distance x. Determine the net force (per unit length) acting on wire 2 and the angular frequency of the resulting oscillation. Assume the mass per unit length of wire 2 is λ and x ≪ d.

FIGURE P30.91

To determine

The net force per unit length acting on wire $2$.

Answer to Problem 91PQ

The net force (per unit length) acting on wire $2$ is $-\left(\frac{{\mu }_{0}I{I}_0}{\pi d^2}\right)x$.

Explanation of Solution

The wire $2$ is displaced to the right by a small distance $x$ as shown in figure below.

    

The direction of magnetic field on wire $2$ due to $1$ and due to $2$ and directions of magnetic force on wire $2$ due to $1$ and due to $2$ are also shown in above figure.

Write the expression of the force between two parallel current carrying conductors per unit length.

    $\frac{F_{ba}}{l}=\frac{{\mu }_0{I}_a{I}_b}{2\pi d}$                                                                                           (I)

Here, $F_{ba}$ is the force on wire $b$ due to wire $a$ , $I_a$ is the current flowing in wire $a$, $I_b$ is the current flowing in wire $b$, $l$ is the length of wire, $d$ is the distance between $a$ and $b$ wire.

Substitute $F_{21}$ for $F_{ba}$, $I_0$ for $I_a$, $I$ for $I_b$, and $\left(d+x\right)$ for $d$ in equation (I).

    $\frac{F_{21}}{l}=\frac{{\mu }_{0}I{I}_0}{2\pi \left(d+x\right)}$                                                                                     (II)

Here, $I_0$ is the current in wire 2 and $F_{21}$ is the force on wire 2 due to wire 1.

Substitute $I_0$ for $I_a$, $I$ for $I_b$ and $\left(d-x\right)$ for $d$ in equation (I).

    $\frac{F_{23}}{l}=\frac{{\mu }_{0}I{I}_0}{2\pi \left(d-x\right)}$                                                                                     (III)

Here, $F_{23}$ is the force on wire 2 due to wire 3.

Write the expression for the net force on wire $2$.

    $\frac{F_{net}}{l}=\frac{F_{21}}{l}-\frac{F_{23}}{l}$                                                                                       (IV)

Conclusion:

Substitute $\frac{{\mu }_{0}I{I}_0}{2\pi \left(d+x\right)}$ for $\frac{F_{21}}{l}$ and $\frac{{\mu }_{0}I{I}_0}{2\pi \left(d-x\right)}$ for $\frac{F_{23}}{l}$ in equation (IV).

    $\begin{array}{c} \frac{F_{net}}{l}=\frac{{\mu }_{0}I{I}_0}{2\pi \left(d+x\right)}-\frac{{\mu }_{0}I{I}_0}{2\pi \left(d-x\right)}\\ =\frac{{\mu }_{0}I{I}_0}{2\pi }\left[\frac{1}{\left(d+x\right)}-\frac{1}{\left(d-x\right)}\right]\\ =\frac{{\mu }_{o}I{I}_o}{2\pi }\frac{\left(-2x\right)}{d^2-x^2}\end{array}$

Since $x<<d$

This implies that $d^2-x^2\approx d^2$

    $\frac{F_{net}}{l}=-\left(\frac{{\mu }_{0}I{I}_0}{\pi d^2}\right)x$                                                                                     (V)

Thus, the net force (per unit length) acting on wire $2$ is $-\left(\frac{{\mu }_{0}I{I}_0}{\pi d^2}\right)x$.

To determine

The angular frequency of the resulting oscillation.

Answer to Problem 91PQ

The angular frequency of the resulting oscillating is $\sqrt{\frac{{\mu }_{0}I{I}_0}{\pi \lambda d^2}}$.

Explanation of Solution

Rewrite the equation (V).

    $\frac{F_{net}}{l}+\left(\frac{{\mu }_{0}I{I}_0}{\pi d^2}\right)x=0$                                                                                   (VI)

Since, the wire $2$ displaced by a small distance $x$,

Write the expression for the force experienced by the wire

    $\frac{F_{net}}{l}=\frac{m}{l}\frac{d^{2}x}{d{t}^2}$

Rearrange the above expression.

  $\frac{F_{net}}{l}=\lambda \frac{d^{2}x}{d{t}^2}$                                                                                           (VII)

Here, $\lambda$ is the mass per unit length and $m$ is the mass of wire.

Substitute $\lambda \frac{d^{2}x}{d{t}^2}$ for $\frac{F_{net}}{l}$ in equation (VI).

    $\lambda \frac{d^{2}x}{d{t}^2}+\left(\frac{{\mu }_{0}I{I}_0}{\pi d^2}\right)x=0$

Rearrange the above expression.

  $\frac{d^{2}x}{d{t}^2}+\left(\frac{{\mu }_{0}I{I}_0}{\pi \lambda d^2}\right)x=0$                                                                                (VIII)

Write the general expression for our oscillating wave.

    $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$                                                                                             (IX)

Here, $\omega$ is the oscillation frequency.

Compare equation (VIII) is and (IX) for the coefficient of $x$ .

    $\begin{array}{l}{\omega }^2=\frac{{\mu }_{0}I{I}_0}{\pi \lambda d^2}\\ \omega =\sqrt{\frac{{\mu }_{0}I{I}_0}{\pi \lambda d^2}}\end{array}$

Conclusion:

Hence, the angular frequency of the resulting oscillation is $\sqrt{\frac{{\mu }_{0}I{I}_0}{\pi \lambda d^2}}$