Chapter 23, Problem 54P

(a)

To determine

The current in the coil

Answer to Problem 54P

The current in the coil has a value of $1.43\times {10}^5\text{A}$.

Explanation of Solution

Write the equation for the magnetic field at the center of the coil.

    $B_{center}=\frac{{\mu }_{0}I}{2a}$        (I)

Here, ${\mu }_0$ is the permeability in free space, $I$ is the current in the coil and $a$ is the radius of the coil. Rearrange equation (I) to find the equation for the current in the coil.

    $I=\frac{2a{B}_{center}}{{\mu }_0}$

Conclusion:

Substitute $6\text{cm}$ for $a$, $1.50\text{T}$ for $B_{center}$ and $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_0$.

    $\begin{array}{c}I=\frac{2\left(0.060\text{m}\right)\left(1.50\text{T}\right)}{4\pi \times {10}^{-7}\text{T}\text{.m/A}}\\ =1.43\times {10}^5\text{A}\end{array}$

Therefore, a current of $1.43\times {10}^5\text{A}$ exists in the coil.

(b)

To determine

The rate of change of magnetic field

Answer to Problem 54P

The rate of change of magnetic field is $82.4\text{T/s}$.

Explanation of Solution

Write the equation for the magnetic field along the axis of the coil at a distance $x$.

    $B_{axis}=\frac{{\mu }_{0}I{a}^2}{2{\left(a^2+x^2\right)}^{3/2}}$        (II)

Here, ${\mu }_0$ is the permeability in free space, $I$ is the current in the coil and $a$ is the radius of the coil.

Differentiate the left hand side and the right hand side of the equation with respect to time $t$.

    $\frac{d{B}_{axis}}{dt}=\frac{{\mu }_0{a}^2}{2{\left(a^2+x^2\right)}^{3/2}}\frac{dI}{dt}$

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_0$, $6\text{cm}$ for $a$, $2.50\text{cm}$ for $x$ and $1.00\times {10}^7\text{A/s}$ for $dI/dt$.

    $\begin{array}{c}\frac{d{B}_{axis}}{dt}=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m/A}\right){\left(0.060\text{m}\right)}^2}{2{\left(0.060{\text{m}}^2+0.0250{\text{m}}^2\right)}^{3/2}}\left(1.00\times {10}^7\text{A/s}\right)\\ =\frac{0.045}{2{(0.0042)}^{3/2}}\text{T/s}\\ \text{=}\text{82}\text{.4}\text{T/s}\end{array}$

Therefore, the rate of change of magnetic field is $82.4\text{T/s}$.

(c)

To determine

The distance from the coil at which the metals experience a magnetic field

Answer to Problem 54P

The metals or devices experience a magnetic field at a distance of $86.3\text{cm}$ from the coil.

Explanation of Solution

Rewrite equation (II) to find the equation for $x$.

    $\begin{array}{c}{B}_{axis}=\frac{{\mu }_{0}I{a}^2}{2{\left(a^2+x^2\right)}^{3/2}}\\ {\left(a^2+x^2\right)}^{3/2}=\frac{{\mu }_{0}I{a}^2}{2B_{axis}}\\ x={\left(\sqrt[3]{{\left(\frac{{\mu }_{0}I{a}^2}{2B_{axis}}\right)}^2}-a^2\right)}^{1/2}\end{array}$

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_0$, $1.43\times {10}^5\text{A}$ for $I$, $6\text{cm}$ for $a$ and $5\times {10}^{-4}\text{T}$ for $B_{axis}$.

    $\begin{array}{c}x={\left(\sqrt[3]{{\left(\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m/A}\right)\left(1.43\times {10}^5\text{A}\right){\left(0.060\text{m}\right)}^2}{2\left(5\times {10}^{-4}\text{T}\right)}\right)}^2}-{\left(0.060\text{m}\right)}^2\right)}^{1/2}\\ =0.863\text{m}\\ =\text{86}\text{.3}\text{cm}\end{array}$

Therefore, the metals or devices experience a magnetic field at a distance of $86.3\text{cm}$ from the coil.