Chapter 31, Problem 11P

(a)

To determine

The induced current in the ring.

Answer to Problem 11P

The induced current in the ring is $1.60\text{A}$ in counterclockwise direction.

Explanation of Solution

Write the expression to calculate the magnetic flux through the loop.

    ${\Phi }_{\text{B}}=\frac{BA}{2}$                                                                                                                  (I)

Here, ${\Phi }_{\text{B}}$ is the magnetic flux, $B$ is the magnetic field, $A$ is the area

Write the expression for the induced emf by Faraday law.

    $E=\frac{d}{dt}\left({\Phi }_{\text{B}}\right)$                                                                                                              (II)

Here, $E$ is the induced emf.

Write the expression for the area.

    $A=\pi r_2{}^2$                                                                                                                   (III)

Here, $r_2$ is the radius of the solenoid.

Write the expression for the magnetic field.

    $B={\mu }_{0}NI$                                                                                                                (IV)

Here, $B$ is the magnetic field, $N$ is the number of turns, $I$ is the current an ${\mu }_0$ is the permeability.

Substitute $\frac{BA}{2}$ for ${\Phi }_{\text{B}}$, $\pi r_2{}^2$ for $A$ and ${\mu }_{0}NI$ for $B$ in equation (II).

    $\begin{array}{c}E=\frac{d}{dt}\left(\frac{\left({\mu }_{0}NI\right)\left(\pi r_2{}^2\right)}{2}\right)\\ =\left(\frac{\left({\mu }_{0}N\right)\left(\pi r_2{}^2\right)}{2}\right)\frac{dI}{dt}\end{array}$                                                                                           (V)

Write the expression for the induced current.

    $i=\frac{E}{R}$                                                                                                                         (VI)

Substitute $\left(\frac{\left({\mu }_{0}N\right)\left(\pi r_2{}^2\right)}{2}\right)\frac{dI}{dt}$ for $E$ in equation (VI),

    $i=\frac{1}{R}\left(\left(\frac{\left({\mu }_{0}N\right)\left(\pi r_2{}^2\right)}{2}\right)\frac{dI}{dt}\right)$                                                                                   (VII)

Conclusion:

Substitute $3.00\times {10}^{-4}\Omega$ for $R$, $3.00\text{cm}$ for $r_2$, $270\text{A}/\text{s}$ for $\frac{dI}{dt}$, $4\pi \times {10}^{-7}\text{T-m}/\text{A}$ for ${\mu }_0$ and $1000\text{turns}/\text{m}$ for $N$ in equation (VII) to solve for $i$.

    $\begin{array}{c}i=\left(\begin{array}{l}\left(\frac{1}{3.00\times {10}^{-4}\Omega }\right)\left(270\text{A}/\text{s}\right)\\ \left(\frac{\left(4\pi \times {10}^{-7}\text{T-m}/\text{A}\right)\left(1000\text{turns}/\text{m}\right)\left(\pi {\left(3.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}{2}\right)\end{array}\right)\\ =1.598\text{A}\\ \approx \text{1}\text{.60}\text{A}\end{array}$

The direction of the current is counter clockwise.

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

(b)

To determine

The magnitude of the magnetic field at the center of the ring

Answer to Problem 11P

The magnitude of the magnetic field at the center of the ring is $20.1\mu \text{T}$.

Explanation of Solution

Write the expression to calculate the magnetic field at the center of the ring.

    $B^{\prime }=\frac{{\mu }_{0}i}{2r_1}$                                                                                                                 (VII)

Here, $B^{\prime }$ is the magnetic field at the center of the ring and $r_1$ is the radius of the ring.

Conclusion:

Substitute $5.00\text{cm}$ for $r_1$, $1.60\text{A}$ for $i$ and $4\pi \times {10}^{-7}\text{T-m}/\text{A}$ for ${\mu }_0$ in equation (VIII) to solve for $B^{\prime }$.

    $\begin{array}{c}{B}^{\prime }=\frac{\left(4\pi \times {10}^{-7}\text{T-m}/\text{A}\right)\left(1.60\text{A}\right)}{2\left(5.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =20.1\times {10}^{-6}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =20.1\mu \text{T}\end{array}$

Therefore, the magnitude of the magnetic field at the center of the ring is $20.1\mu \text{T}$.

(c)

To determine

The direction of the magnetic field at the center of the ring.

Answer to Problem 11P

The direction of the magnetic field at the center of the ring is to the left.

Explanation of Solution

The magnetic field of the solenoid points towards the right but the induced field opposes the original field. So, the magnetic field at the center of the ring will be directed towards the left.

Conclusion:

Therefore, the direction of the magnetic field at the center of the ring is to the left.