Chapter 14, Problem 1SP

(a)

To determine

The magnitude of magnetic force per unit length exerted by one wire on the other.

Answer to Problem 1SP

The force is $2.0\times {10}^{-4}\text{N}/\text{m}$.

Explanation of Solution

Given Info The current through the wires are $5\text{A}$ and $10\text{A}$, the separation between wires is $5\text{cm}$.

Write the formula to calculate the magnetic force per unit length between two parallel current carrying wires.

$\frac{F}{l}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}$

Here,

$F$ is the magnetic force exerted by one on other

$l$ is the length of wire on which force is exerted

${\mu }_0$ is the permeability of free space

$I_1$ is the current through first wire

$I_2$ is the current through second wire

$d$ is the separation between wires

The value of ${\mu }_0$ is $4\pi \times {10}^{-7}\text{H}$.

Substitute $4\pi \times {10}^{-7}\text{H}$ for ${\mu }_0$,$5\text{A}$ for $I_1$,$10\text{A}$ for $I_2$,$5\text{cm}$ for $d$ and $1\text{m}$ for $l$ in the above equation to find $F$.

$\begin{array}{c}\frac{F}{1\text{m}}=\frac{\left(4\pi \times {10}^{-7}\text{H}\right)}{2\pi }\frac{\left(5\text{A}\right)\left(10\text{A}\right)}{5\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =2.0\times {10}^{-4}\text{N}/\text{m}\end{array}$

Conclusion:

Therefore, the force is $2.0\times {10}^{-4}\text{N}/\text{m}$.

(b)

To determine

The direction of force exerted by one on the other wire.

Answer to Problem 1SP

The force will be perpendicular to both current and magnetic field.

Explanation of Solution

The direction of magnetic force exerted by one on other can be found using right hand thumb rule. Use right hand to find the direction. Let the index finger point in direction of current flow, middle finger points in direction of magnetic field, direction of magnetic force is given by the right hand thumb.

From the above rule, it is clear that the force acts in perpendicular direction to both current and magnetic field. Since the current flow through wires are in opposite directions, the wires will repel each other.

Conclusion:

Therefore, the force will be perpendicular to both current and magnetic field.

(c)

To determine

The magnetic force exerted on wire with current $10\text{A}$ on its30 cm length.

Answer to Problem 1SP

The force is $6.0\times {10}^{-5}\text{N}$.

Explanation of Solution

Given Info The current through the wires are $5\text{A}$ and $10\text{A}$, the separation between wires is $5\text{cm}$.

Write the formula to calculate the magnetic force between two parallel current carrying wires.

$F=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_{2}l}{d}$

Substitute $4\pi \times {10}^{-7}\text{H}$ for ${\mu }_0$,$5\text{A}$ for $I_1$,$10\text{A}$ for $I_2$ ,$5\text{cm}$ for $d$ and $30\text{cm}$ for $l$ in the above equation to find $F$.

$\begin{array}{c}F=\frac{\left(4\pi \times {10}^{-7}\text{H}\right)}{2\pi }\frac{\left(5\text{A}\right)\left(10\text{A}\right)\left(\text{30}\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}{5\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =6.0\times {10}^{-5}\text{N}\end{array}$

Conclusion:

Therefore, the force is $6.0\times {10}^{-5}\text{N}$.

(d)

To determine

The magnetic field produced by $5\text{A}$ wire at the position of $10\text{A}$ wire.

Answer to Problem 1SP

The magnetic field is $2.0\times {10}^{-5}\text{T}$.

Explanation of Solution

Given Info: The separation between wires is $5\text{cm}$.

Write the formula to calculate magnetic field from magnetic force.

$B=\frac{F}{I_{2}l}$

Here,

$B$ is the magnitude of the magnetic field

Substitute $6.0\times {10}^{-5}\text{N}$ for $F$,$10\text{A}$ for $I_2$ , and $30\text{cm}$ for $l$ in the above equation to find $F$.

$\begin{array}{c}B=\frac{6.0\times {10}^{-5}\text{N}}{\left(10\text{A}\right)\left(30\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}\\ =2.0\times {10}^{-5}\text{T}\end{array}$

Conclusion:

Therefore, the magnetic field is $2.0\times {10}^{-5}\text{T}$.

(e)

To determine

The direction of magnetic field produced by $5\text{A}$ wire at the position of $10\text{A}$ wire.

Answer to Problem 1SP

The magnetic field is perpendicular to plane of the page and points into the page.

Explanation of Solution

The direction of magnetic field due to current flow is given by right hand thumb rule. If right hand’s thumb points in direction of current flow, curled fingers represent the direction of magnetic field.

In the figure, the current $10\text{A}$ flows in the downward direction. Pointing the right thumb in the downward direction makes the fingers curl into the page. Thus, magnetic field produced by $5\text{A}$ wire at the position of $10\text{A}$ wire points into the page.

Conclusion:

Thus, the magnetic field is perpendicular to plane of the page and points into the page.