Chapter 22, Problem 42P

(a)

To determine

The possibility of the situation and any other possibility of the situation.

Answer to Problem 42P

Yes, the situation is possible and the only possibility is placing the third wire at a distance at which the force between wire 1 and 3 balances the force between wire 2 and 3.

Explanation of Solution

Given info: The electric current in wire 1 is $1.50\text{A}$ in upward direction and the electric current in wire 2 is $4.00\text{A}$ in downward direction. The distance between wire 1 and wire 2 is $20.0\text{cm}$

The orientation of the wires is shown in the figure below.

Figure (1)

The formula to calculate the magnitude of force per unit length is,

$F=L\frac{{\mu }_{\circ }{I}_1{I}_2}{2\pi d}$ (1)

Here,

$I_1$ is the electric current in wire 1.

$I_2$ is the electric current in wire 2.

$d$ is the distance between the wires.

${\mu }_{\circ }$ is the permeability of free space.

$L$ is the length of wire on which force is calculated.

For the stability of wire 3 the magnetic force due to wire 1 on 3 must be equal to that of magnetic force due to wire 2 on 3

For equilibrium of wire 3,

$F_{\text{1}\text{on}\text{3}}=F_{2\text{on}\text{3}}$

From equation (1)

$L\frac{{\mu }_{\circ }{I}_1{I}_3}{2\pi d}=L\frac{{\mu }_{\circ }{I}_3{I}_2}{2\pi \left(x+d\right)}$

The above equation would be satisfied for only a particular position of wire 3 with respect to wire 1 and 2.

Thus, the situation is possible and the only possibility is placing the third wire at a distance at which the force between wire 1 and 3 balances the force between wire 2 and 3.

Conclusion:

Therefore, the situation is possible and the only possibility is placing the third wire at a distance at which the force between wire 1 and 3 balances the force between wire 2 and 3.

(b)

To determine

The position of wire 3.

Answer to Problem 42P

The position of wire 3 is $12\text{cm}$ on the left of wire 1.

Explanation of Solution

Given info: The electric current in wire 1 is $1.50\text{A}$ in upward direction and the electric current in wire 2 is $4.00\text{A}$ in downward direction. The distance between wire 1 and wire 2 is $20.0\text{cm}$

The formula to calculate the magnitude of force per unit length is,

$F=L\frac{{\mu }_{\circ }{I}_1{I}_2}{2\pi d}$ (2)

Here,

$I_1$ is the electric current in wire 1.

$I_2$ is the electric current in wire 2.

$d$ is the distance between the wires.

${\mu }_{\circ }$ is the permeability of free space.

For the stability of wire 3 the magnetic force due to wire 1 on 3 must be equal to that of magnetic force due to wire 2 on 3

For equilibrium of wire 3,

$F_{\text{1}\text{on}\text{3}}=F_{2\text{on}\text{3}}$

From equation (2)

$\begin{array}{c}L\frac{{\mu }_{\circ }{I}_1{I}_3}{2\pi d}=L\frac{{\mu }_{\circ }{I}_3{I}_2}{2\pi \left(x+d\right)}\\ \frac{I_1}{d}=\frac{I_2}{\left(x+d\right)}\end{array}$

Substitute $1.50\text{A}$ for $I_1$, $4.00\text{A}$ for $I_2$ and $20.0\text{cm}$ for $x$ in the above equation to find the value of $d$.

$\begin{array}{c}\frac{\left(1.50\text{A}\right)}{d}=\frac{4.00\text{A}}{\left(20\text{cm}+d\right)}\\ 4.0d=1.50\left(20\text{cm}+d\right)\\ d=12\text{cm}\end{array}$

The distance of the wire 3 is $12\text{cm}$ on the left of wire 1.

Thus, the position of wire 3 is $12\text{cm}$ on the left of wire 1.

Conclusion:

Therefore, the position of wire 3 is $12\text{cm}$ on the left of wire 1.

(c)

To determine

The magnitude and direction of the current in wire 3.

Answer to Problem 42P

The magnitude of current in wire 3 is $2.4\text{A}$ in downward direction.

Explanation of Solution

Given info: The electric current in wire 1 is $1.50\text{A}$ in upward direction and the electric current in wire 2 is $4.00\text{A}$ in downward direction. The distance between wire 1 and wire 2 is $20.0\text{cm}$

For the stability of wire 1 the magnetic force due to wire 1 on 3 must be equal to that of magnetic force due to wire 2 on 1

For equilibrium of wire 3,

$F_{3\text{on}1}=F_{2\text{on}1}$

From equation (2)

$\begin{array}{c}L\frac{{\mu }_{\circ }{I}_1{I}_3}{2\pi d}=L\frac{{\mu }_{\circ }{I}_1{I}_2}{2\pi \left(x+d\right)}\\ \frac{I_3}{d_1}=\frac{I_2}{d_2}\end{array}$

Here,

$d_1$ is the distance between wire 1 and 3.

$d_2$ is the distance between wire 1 and 2.

Substitute $4.00\text{A}$ for $I_2$, $12\text{cm}$ for $d_1$ and $20.0\text{cm}$ for $d_2$ in the above equation to find the value of $I_3$.

$\begin{array}{c}\frac{I_3}{\left(12\text{cm}\right)}=\frac{4.00\text{A}}{\left(20.0\text{cm}\right)}\\ I_3=2.4\text{A}\end{array}$

The magnitude of electric current in wire 3 is $2.4\text{A}$ in downward direction as the current is in the same direction as that of the wire 2.

Conclusion:

Therefore, the magnitude of current in wire 3 is $2.4\text{A}$ in downward direction.