Chapter 30, Problem 29PQ

(a)

To determine

Find the magnitude and direction of the magnetic field due to the two loops at the origin.

Answer to Problem 29PQ

The direction of the magnetic field is out of the page and magnitude of magnetic field is $\underset{\_}{1.02\times {10}^{-5}\text{ T}}$.

Explanation of Solution

Assume that magnetic field due to smaller loop is $B_1$ and magnetic field due to larger loop is $B_2$.

Write the expression for magnetic field due to smaller loop as.

    $B_1=\frac{{\mu }_0}{4\pi }\left(\frac{2\pi i_1}{r_1}\right)$                                                                                             (I)

Here, $i_1$ is the current, $r_1$ is the radius and ${\mu }_0$ is the permeability of smaller loop. The direction of the magnetic field is inward.

Write the expression for magnetic field due to larger loop as.

    $B_2=\frac{{\mu }_0}{4\pi }\left(\frac{2\pi i_2}{r_2}\right)$                                                                                           (II)

Here, $i_2$ is the current, $r_2$ is the radius and ${\mu }_0$ is the permeability of larger loop.

The direction of the magnetic field is outward.

Write the expression for net magnetic field as.

    $B_{net}=B_2-B_1$

Substitute $\frac{{\mu }_0}{4\pi }\left(\frac{2\pi i_2}{r_2}\right)$ for $B_2$ and $\frac{{\mu }_0}{4\pi }\left(\frac{2\pi i_1}{r_1}\right)$ for $B_1$ in above equation.

$\begin{array}{c}=\left(\frac{{\mu }_0}{4\pi }\right)2\pi \left[\frac{i_2}{r_2}-\frac{i_1}{r_1}\right]\\ =\frac{{\mu }_0\times 2\pi }{4\pi }\left[\frac{i_2}{r_2}-\frac{i_1}{r_1}\right]\end{array}$

Rearrange the above equation as.

    $B_{net}=\frac{{\mu }_0}{2}\left[\frac{i_2}{r_2}-\frac{i_1}{r_1}\right]$                                                                                       (III)

Conclusion:

Substitute $4.00\text{ A}$ for $i_1$, $9.00\text{ A}$ for $i_2$, $10.0\text{ cm}$ for $r_1$, $16.0\text{ cm}$ for $r_2$ and $4\pi \times {10}^{-7}{\text{ N/A}}^2$ for ${\mu }_0$ in equation (III).

    $\begin{array}{c}{B}_{net}=\frac{4\pi \times {10}^{-7}\left({\text{N/A}}^{\text{2}}\right)}{2}\left[\frac{9.00\text{ A}}{16.0\times {10}^{-2}\text{ m}}-\frac{4.00\text{ A}}{10.0\times {10}^{-2}\text{ m}}\right]\\ =1.02\times {10}^{-5}\text{ T}\end{array}$

Thus the direction of the magnetic field is out of the page and magnitude of magnetic field is $\underset{\_}{1.02\times {10}^{-5}\text{ T}}$.

(b)

To determine

Determine the radius $r_2$ for the magnetic field at the origin to be zero.

Answer to Problem 29PQ

The magnetic field at the origin to be zero for $r_2$ is $\underset{\_}{0.225\text{ m}}$

Explanation of Solution

When $r_1$ is held constant at 10.0 cm, then magnetic field of smaller loop will be equal to magnetic field of larger loop.

Write the expression for the magnetic field at the origin to be zero as.

    $\begin{array}{c}{B}_1=B_2\\ \frac{{\mu }_0}{4\pi }\left(\frac{2\pi i_1}{r_1}\right)=\frac{{\mu }_0}{4\pi }\left(\frac{2\pi i_2}{r_2}\right)\end{array}$

Rearrange the above equation as.

    $\frac{i_1}{r_1}=\frac{i_2}{r_2}$

Rearrange the above equation as.

    $r_2=\frac{i_2\times r_1}{i_1}$                                                                                                    (IV)

Here, $i_1$ is the current, $r_1$ is the radius $i_2$ is the current and $r_2$ is the radius.

Conclusion:

Substitute $4.00\text{ A}$ for $i_1$, $9.00\text{ A}$ for $i_2$ and $10.0\text{ cm}$ for $r_1$ in equation (IV).

    $\begin{array}{c}{r}_2==\frac{9.00\text{A}\times 10.0\text{cm}}{4.00\text{A}}\\ =\frac{9.00\text{A}\times 10.0\times {10}^{-2}\text{m}}{4.00\text{A}}\\ =0.225\text{ m}\end{array}$

Thus, the magnetic field at the origin to be zero for $r_2$ is $\underset{\_}{0.225\text{ m}}$.