Chapter 23, Problem 12P

A piece of insulated wire is shaped into a figure eight as shown in Figure P23.12. For simplicity, model the two halves of the figure eight as circles. The radius of the upper circle is 5.00 cm and that of the lower circle is 9.00 cm. The wire has a uniform resistance per unit length of 3.00 Ω/m. A uniform magnetic field is applied perpendicular to the plane of the two circles, in the direction shown. The magnetic field is increasing at a constant rate of 2.00 T/s. Find (a) the magnitude and (b) the direction of the induced current in the wire.

Figure P23.12

To determine

The magnitude of the induced current

Answer to Problem 12P

The induced current has a magnitude of $0.0133\text{A}$.

Explanation of Solution

The magnetic field in which the wire is placed is increasing and points into the page. This increasing flux is opposed by a magnetic field out of the page due to the counterclockwise current as a result of induced emf in the upper loop. Similarly, the lower loop also develops an induced emf.

The emf generated in the upper and the lower loops are in opposite directions as the two loops cross over. Write the equation for the net emf generated which is the difference between the emf generated in the two loops.

    ${\epsilon }_{net}={\epsilon }_2-{\epsilon }_1$        (I)

Here, ${\epsilon }_{net}$ is the net emf generated, ${\epsilon }_2$ is the emf in the lower loop and ${\epsilon }_1$ is the emf in the lower loop. Substitute $A_2\left(dB/dt\right)$ for ${\epsilon }_2$ and $A_1\left(dB/dt\right)$ for ${\epsilon }_1$ in equation (I).

    ${\epsilon }_{net}=A_2\left(\frac{dB}{dt}\right)-A_1\left(\frac{dB}{dt}\right)$        (II)

Here, $A_2$ is the area of the lower loop, $A_1$ is the area of the upper loop and $\left(dB/dt\right)$ is the change in magnetic field. Substitute $\pi r_2^2$ for $A_2$ and $\pi r_1^2$ for $A_1$ in equation (II).

    $\begin{array}{c}{\epsilon }_{net}=\pi r_2^2\left(\frac{dB}{dt}\right)-\pi r_1^2\left(\frac{dB}{dt}\right)\\ =\pi \frac{dB}{dt}\left(r_2^2-r_1^2\right)\end{array}$        (III)

Here, $r_2$ is the radius of the lower loop and $r_1$ is the radius of the upper loop. Write the equation for the induced current in the wire.

    $I=\frac{{\epsilon }_{net}}{R}$        (IV)

Here, $I$ is the current induced in the wire, ${\epsilon }_{net}$ is the net emf generated and $R$ is the resistance of the wire. Substitute equation (III) in equation (IV).

    $I=\frac{\pi \frac{dB}{dt}\left(r_2^2-r_1^2\right)}{R}$        (V)

Multiply and divide the denominator of equation (V) with the length of the wire $l$.

    $I=\frac{\pi \frac{dB}{dt}\left(r_2^2-r_1^2\right)}{\left(\frac{R}{l}\right)l}$        (VI)

The length of the wire is the sum of the circumferences of the lower loop and the upper loop. The circumference of the lower loop is $2\pi r_2$ and the circumference of the upper loop is $2\pi r_1$. Use this to rewrite the equation (VI) and solve.

    $\begin{array}{c}I=\frac{\pi \frac{dB}{dt}\left(r_2^2-r_1^2\right)}{\left(\frac{R}{l}\right)\left(2\pi r_2+2\pi r_1\right)}\\ =\frac{\pi \frac{dB}{dt}\left(r_2-r_1\right)\left(r_2+r_1\right)}{2\pi \left(\frac{R}{l}\right)\left(r_2+r_1\right)}\\ =\frac{\frac{dB}{dt}\left(r_2-r_1\right)}{2\left(\frac{R}{l}\right)}\end{array}$

Conclusion:

Substitute $2.00\text{T/s}$ for $dB/dt$, $9.00\text{cm}$ for $r_2$, $5.00\text{cm}$ for $r_1$ and $3.00\Omega \text{/m}$ for $R/l$.

    $\begin{array}{c}I=\frac{2.00\text{T/s}\left(9.00\text{cm}-5.00\text{cm}\right)}{2\left(3.00\Omega \text{/m}\right)}\\ =0.0133\text{A}\end{array}$

Therefore, the induced current has a magnitude of $0.0133\text{A}$.

To determine

The direction of the induced current

Answer to Problem 12P

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

Explanation of Solution

The lower loop has a larger area and hence the emf. As a result, the change in magnetic flux is also larger in the lower loop.

This increasing flux is opposed by a magnetic field out of the page due to the counterclockwise current as a result of induced emf in the upper loop.

Conclusion:

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