Chapter 31, Problem 31.20P

(a)

To determine

The magnitude of the induced current in the wire.

Answer to Problem 31.20P

The magnitude of the induced current in the wire is $0.0133\text{A}$.

Explanation of Solution

Given info: The radius of the upper circle is $5.00\text{cm}$, the radius of the lower circle is $9.00\text{cm}$, the value of the resistance per unit length is $3.00\text{Ω}/\text{m}$, the angle between the magnetic field and the normal to the loop is $0°$ and the rate of the magnetic field is $2.00\text{T}/\text{s}$.

Write the expression for the area of the loop.

$A=\pi r^2$ (1)

Here,

$r$ is the radius of the loop.

For upper circle.

Substitute $5.00\text{cm}$ for $r$ in equation (1).

$\begin{array}{c}{A}_{\text{upper}}=\pi {\left(5.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ \approx 7.85\times {10}^{-3}{\text{m}}^2\end{array}$

Thus, the area of the upper circle is $7.85\times {10}^{-3}{\text{m}}^2$.

For lower circle.

Substitute $9.00\text{cm}$ for $r$ in equation (1).

$\begin{array}{c}{A}_{\text{lower}}=\pi {\left(9.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ \approx 25.45\times {10}^{-3}{\text{m}}^2\end{array}$

Thus, the area of the lower circle is $25.45\times {10}^{-3}{\text{m}}^2$.

Write the expression for the magnetic flux through a coil

${\varphi }_{\text{B}}=BA\cos \theta$

Here,

$A$ is the area of the loop.

$B$ is the magnitude of the magnetic field.

$\theta$ is the angle between the magnetic field and the normal to the loop.

Write the expression for the emf of the coil.

$\epsilon =-N\frac{d{\varphi }_{\text{B}}}{dt}$

Here,

$N$ is the number of turns.

$t$ is the time.

Substitute $BA\cos \theta$ for $\varphi$ in the above equation.

$\begin{array}{c}\epsilon =-\frac{d\left(BA\cos \theta \right)}{dt}\\ =-\frac{A\cos \theta \left(dB\right)}{dt}\end{array}$ (2)

For the upper circle.

Substitute ${\epsilon }_{\text{upper}}$ for $\epsilon$, $0°$ for $\theta$, $7.85\times {10}^{-3}{\text{m}}^2$ for $A$ and $2.00\text{T}/\text{s}$ for $\frac{dB}{dt}$ in the equation (2).

$\begin{array}{c}{\epsilon }_{\text{upper}}=-\left(1\right)\left(7.85\times {10}^{-3}{\text{m}}^2\right)\left(\cos 0°\right)\left(2.00\text{T}/\text{s}\right)\\ =-1.57\times {10}^{-2}\text{V}\end{array}$

Thus, the emf induced in the upper loop is $-1.57\times {10}^{-2}\text{V}$.

For the lower circle.

Substitute ${\epsilon }_{\text{lower}}$ for $\epsilon$, $180°$ for $\theta$, $25.45\times {10}^{-3}{\text{m}}^2$ for $A$ and $2.00\text{T}/\text{s}$ for $\frac{dB}{dt}$ in the equation (2).

$\begin{array}{c}{\epsilon }_{\text{lower}}=-\left(1\right)\left(25.45\times {10}^{-3}{\text{m}}^2\right)\left(\cos 180°\right)\left(2.00\text{T}/\text{s}\right)\\ =5.09\times {10}^{-2}\text{V}\end{array}$

Thus, the emf induced in the upper loop is $5.09\times {10}^{-2}\text{V}$.

Write the expression for the resistance.

$R_{\text{net}}=R\left(2\pi r\right)$ (3)

Here,

$R$ is the resistance per unit length in the loop.

For upper circle.

Substitute $5.00\text{cm}$ for $r$ and $3.00\text{Ω}/\text{m}$ for $R$ in equation (3).

$\begin{array}{c}{R}_{\text{upper}}=\left(3.00\text{Ω}/\text{m}\right)\left(2\pi \times 5.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ \approx 0.94\text{Ω}\end{array}$

Thus, the resistance in the upper circle is $0.94\text{Ω}$.

For lower circle.

Substitute $9.00\text{cm}$ for $r$ and $3.00\text{Ω}/\text{m}$ for $R$ in equation (3).

$\begin{array}{c}{R}_{\text{lower}}=\left(3.00\text{Ω}/\text{m}\right)\left(2\pi \times 9.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ \approx 1.696\text{Ω}\end{array}$

Thus, the resistance in the lower circle is $1.696\text{Ω}$.

Write the expression for the induced current.

$I=\frac{{\epsilon }_{\text{lower}}-{\epsilon }_{\text{upper}}}{R_{\text{lower}}+R_{\text{upper}}}$

Substitute $0.94\text{Ω}$ for $R_{\text{upper}}$, $1.696\text{Ω}$ for $R_{\text{lower}}$, $5.09\times {10}^{-2}\text{V}$ for ${\epsilon }_{\text{lower}}$ and $1.57\times {10}^{-2}\text{V}$ for ${\epsilon }_{\text{upper}}$ in the above equation.

$\begin{array}{c}I=\frac{5.09\times {10}^{-2}\text{V}-1.57\times {10}^{-2}\text{V}}{0.94\text{Ω}+1.696\text{Ω}}\\ =0.0133\text{A}\end{array}$

Conclusion:

Therefore, the magnitude of the induced current in the wire is $0.0133\text{A}$.

(b)

To determine

The direction of the induced current in the wire.

Answer to Problem 31.20P

The current is clockwise in the upper loop and the current is counterclockwise in the lower loop

Explanation of Solution

Given info: The radius of the upper circle is $5.00\text{cm}$, the radius of the lower circle is $9.00\text{cm}$, the value of the resistance per unit length is $3.00\text{Ω}/\text{m}$, the angle between the magnetic field and the normal to the loop is $0°$ and the rate of the magnetic field is $2.00\text{T}/\text{s}$.

Write the expression for the induced current in the wire.

$I=\frac{\epsilon }{R}$

Consider negative current as clockwise and positive current as counterclockwise.

From the above expression, the induced current is directly proportional to the induced emf. From part (a), the value of the induced emf for upper loop is negative, so the value of the current is also negative. The negative current shows the current is clockwise in the upper loop. And also from part (a), the value of the induced emf for lower loop is positive, so the value of the current is also positive. The positive current shows the current is counterclockwise in the lower loop. Thus, the current is clockwise in the upper loop and the current is counterclockwise in the lower loop.

Conclusion:

Therefore, the current is clockwise in the upper loop and the current is counterclockwise in the lower loop.