Chapter 23, Problem 5P

(a)

To determine

The current induced in the aluminum ring

Answer to Problem 5P

A current of $1.60\text{A}$ is induced in the ring.

Explanation of Solution

Write the equation for the emf generated in the coil.

    $\epsilon =-\frac{d{\varphi }_B}{dt}$        (I)

Here, ${\varphi }_B$ is the magnetic flux through the coil. Write the equation for the magnetic flux through the coil.

    ${\varphi }_B=B.A$        (II)

Here, $B$ is the magnetic field due to the solenoid and $A$ is the area of the coil. Substitute equation (IV) in equation (V).

    $\epsilon =-\left(\frac{dBA}{dt}\right)$        (III)

Write the equation for the magnetic field due to the solenoid.

    $B={\mu }_{0}nI$        (IV)

Here, ${\mu }_0$ is the permeability of free space, $n$ is the number of turns in the coil and $I$ is the current in the coil. Substitute equation (IV) in equation (III).

    $\begin{array}{c}\epsilon =-\left(\frac{d{\mu }_{0}nIA}{dt}\right)\\ =-{\mu }_{0}n\frac{dI}{dt}A\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}{\text{.m/A}}^{\text{2}}$ for ${\mu }_0$, $1000\text{/m}$ for $n$, $270\text{A/s}$ for $dI/dt$.

    $\epsilon =-\left(4\pi \times {10}^{-7}\text{T}{\text{.m/A}}^{\text{2}}\right)\left(1000\text{/m}\right)\left(270\text{A/s}\right)A$        (V)

Write the equation for the area of the coil.

    $A=\pi r_2{}^2$

Here, $r_2$ is the radius of the coil. Substitute $0.03\text{m}$ for $r_2$.

    $\begin{array}{c}A=\pi {\left(0.03\text{m}\right)}^2\\ =2.83\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$        (VI)

Substitute equation (VI) in equation (V).

    $\begin{array}{c}\epsilon =-\left(4\pi \times {10}^{-7}\text{T}{\text{.m/A}}^{\text{2}}\right)\left(1000\text{/m}\right)\left(270\text{A/s}\right)\left(2.83\times {10}^{-3}{\text{m}}^{\text{2}}\right)\\ =9.6\times {10}^{-4}\text{V}\end{array}$

The ring is placed at one end of a solenoid. The field in the end of the solenoid is half the field at the center of the solenoid. Write the equation for the emf induced in the ring.

    ${\epsilon }_{ring}=\frac{\epsilon }{2}$

Substitute $9.6\times {10}^{-4}\text{V}$ for $\epsilon$.

    $\begin{array}{c}{\epsilon }_{ring}=\frac{9.6\times {10}^{-4}\text{V}}{2}\\ =4.8\times {10}^{-4}\text{V}\end{array}$

Write the equation for the current induced in the ring.

    $I_{ring}=\frac{{\epsilon }_{ring}}{R}$

Conclusion:

Substitute $4.8\times {10}^{-4}\text{V}$ for ${\epsilon }_0$ and $3.00\times {10}^{-4}\Omega$ for $R$.

    $\begin{array}{c}{I}_{ring}=\frac{4.8\times {10}^{-4}\text{V}}{3.00\times {10}^{-4}\Omega }\\ =1.60\text{A}\end{array}$

Therefore, the current induced in the ring is $1.60\text{A}$.

(b)

To determine

The magnitude of magnetic field in the ring

Answer to Problem 5P

The induced current produces a magnetic field of $20.1\text{μT}$.

Explanation of Solution

Write the equation for the magnetic field produced in the ring.

    $B_{ring}=\frac{{\mu }_0{I}_{ring}}{2r_1}$

Here, ${\mu }_0$ is the permeability of free space, $I_{ring}$ is the current induced in the ring and $r_1$ is the radius of the ring.

Conclusion:

Substitute  $4\pi \times {10}^{-7}\text{T}{\text{.m/A}}^{\text{2}}$ for ${\mu }_0$, $1.60\text{A}$ for $I_{ring}$ and $0.050\text{m}$ for $r_1$.

    $\begin{array}{c}{B}_{ring}=\frac{\left(4\pi \times {10}^{-7}\text{T}{\text{.m/A}}^{\text{2}}\right)\left(1.60\text{A}\right)}{2\left(0.050\text{m}\right)}\\ =2.01\times {10}^{-5}\text{T}\\ \text{=}\text{20}\text{.1}\text{μT}\end{array}$

Therefore, the induced current produces a magnetic field of $20.1\text{μT}$ in the ring.

(c)

To determine

The direction of magnetic field in the ring

Answer to Problem 5P

The magnetic field in the ring points towards the left

Explanation of Solution

Figure (I) shows the direction of the magnetic field in the solenoid.

The magnetic field of the solenoid points to the right as shown in figure.1. Therefore, the magnetic field at the center of the ring acts towards the left in order to oppose the increasing field.

Conclusion:

Therefore, the magnetic field in the ring acts towards the left.