Chapter 21, Problem 5P

An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. Theloop is rotated so that its plane is parallel to the field direction in 0.20 s. What is the average induced emf in the loop?

To determine

The average induced EMF.

Answer to Problem 5P

Solution:

The average induced EMF is equal to 0.20 V.

Explanation of Solution

Initially, the loop of wire is perpendicular to the magnetic field. Therefore, the angle between the magnetic field and a line perpendicular to the face of the loop is taken to be zero degrees.

Finally, the loop is rotated and the plane parallel to the field. Therefore, the angle between magnetic field and the line parallel to the face of loop is 90 degrees.

Given:

The magnetic field of the loop is 1.5 T.

The time taken to rotate the loop is 0.20 s.

The diameter of the loop is 18.5 cm.

Formula Used:

The area of the circular loop is,

$A=\pi r^2$

Here, r is the radius of loop.

The relation between the radius and the diameter is as follows:

$r=\frac{d}{2}$

Here, d is the diameter.

The magnetic flux through the circular lop is as follows:

$\varphi =BA\cos \theta$

Here, B is the magnetic field, A is the area, and $\theta$ is the angle between the area and the magnetic field vector.

The expression for the EMF is as follows:

$\epsilon =-\frac{d\varphi }{dt}$

Here, $d\varphi$ is the change in the flux and $dt$ is the change in time.

Therefore, the equation $\epsilon =-\frac{d\varphi }{dt}$ is rewritten as follows:

$\epsilon =-\frac{{\varphi }_2-{\varphi }_1}{dt}$

Here, ${\varphi }_2$ is the flux in the final position and ${\varphi }_1$ is the flux in the initial situation.

Therefore,

${\varphi }_1=BA\cos {\theta }_{\text{i}}$

Here, ${\theta }_{\text{i}}$ is the initial angle.

And,

${\varphi }_2=BA\cos {\theta }_{\text{f}}$

Here, ${\theta }_{\text{f}}$ is the final angle.

Calculation:

Calculate the radius of the loop by substituting 18.5 cm for d in the equation $r=\frac{d}{2}$.

$\begin{array}{c}r=\frac{18.5\text{ cm}}{2}\\ =\frac{18.5\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}{2}\\ =0.092\text{ m}\end{array}$

Now, substitute $0.092\text{ m}$ for r in the equation $A=\pi r^2$.

$\begin{array}{c}A=\pi {\left(0.092\text{ m}\right)}^2\\ =\left(3.14\right){\left(0.092\text{ m}\right)}^2\\ =0.026{\text{ m}}^2\end{array}$

Now, calculate the magnetic flux in the initial condition by substituting 1.5 T for B, $0.026{\text{ m}}^2$ for A, and $0°$ for ${\theta }_{\text{i}}$ in the equation ${\varphi }_1=BA\cos {\theta }_{\text{i}}$.

$\begin{array}{c}{\varphi }_1=\left(1.5\text{ T}\right)0.026{\text{ m}}^2\cos 0°\\ =0.040\text{ Wb}\end{array}$

And, the magnetic flux in the final condition is obtained by substituting 1.5 T for B, $0.026{\text{ m}}^2$ for A, and $90°$ for ${\theta }_{\text{f}}$ in the equation ${\varphi }_2=BA\cos {\theta }_{\text{f}}$.

$\begin{array}{c}{\varphi }_2=\left(1.5\text{ T}\right)0.026{\text{ m}}^2\cos 90°\\ =0.00\text{ Wb}\end{array}$

Therefore, substitute $0.00\text{ Wb}$ for ${\varphi }_2$, $0.040\text{ Wb}$ for ${\varphi }_1$, and 0.20 s for $dt$ in the equation $\epsilon =-\frac{{\varphi }_2-{\varphi }_1}{dt}$.

$\begin{array}{c}\epsilon =-\frac{0.040\text{ Wb}-0.00\text{ Wb}}{0.20\text{ s}}\\ =-\frac{0.040\text{ Wb}}{0.20\text{ s}}\\ =-0.20\text{ Wb/s}\end{array}$