Chapter 21, Problem 6P

A fixed 10.8-cm-diameter wire coil is perpendicular to a magnetic field 0.48 T pointing up. In 0.16 s,the field is changed to 0.25 T pointing down. What is the average induced emf in the coil?

To determine

The average induced EMF.

Answer to Problem 6P

Solution:

The average induced EMF is equal to $41.75\text{ mV}$.

Explanation of Solution

Given:

The magnetic field of the loop is 0.48 T.

The time taken to rotate the loop is 0.16 s.

The field is changed to 0.25 T.

The diameter of the loop is 10.8 cm.

Formula used:

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 10.8 cm for d in the equation $r=\frac{d}{2}$.

$\begin{array}{c}r=\frac{10.8\text{ cm}}{2}\\ =\frac{10.8\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}{2}\\ =0.054\text{ m}\end{array}$

Now, substitute $0.054\text{ m}$ for r in the equation $A=\pi r^2$.

$\begin{array}{c}A=\pi {\left(0.054\text{ m}\right)}^2\\ =\left(3.14\right){\left(0.054\text{ m}\right)}^2\\ =9.16\times {10}^{-3}{\text{ m}}^2\end{array}$

Now, calculate the magnetic flux in the initial condition by substituting 0.48 T for B, $9.16\times {10}^{-3}{\text{ m}}^2$ for A, and $0°$ for ${\theta }_{\text{i}}$ in the equation ${\varphi }_1=BA\cos {\theta }_{\text{i}}$.

$\begin{array}{c}{\varphi }_1=\left(0.48\text{ T}\right)\left(9.16\times {10}^{-3}{\text{ m}}^2\right)\cos 0°\\ =4.39\times {10}^{-3}\text{ Wb}\end{array}$

And, the magnetic flux in the final condition is obtained by substituting $-0.25\text{ T}$ for B, $9.16\times {10}^{-3}{\text{ m}}^2$ for A, and $90°$ for ${\theta }_{\text{f}}$ in the equation ${\varphi }_2=BA\cos {\theta }_{\text{f}}$.

$\begin{array}{c}{\varphi }_2=\left(-0.25\text{ T}\right)\left(9.16\times {10}^{-3}{\text{ m}}^2\right)\cos 90°\\ =-2.29\times {10}^{-3}\text{ Wb}\end{array}$

Therefore, substitute $-2.29\times {10}^{-3}\text{ Wb}$ for ${\varphi }_2$, $4.39\times {10}^{-3}\text{ Wb}$ for ${\varphi }_1$, and 0.16 s for $dt$ in the equation $\epsilon =-\frac{{\varphi }_2-{\varphi }_1}{dt}$.

$\begin{array}{c}\epsilon =-\frac{4.39\times {10}^{-3}\text{ Wb}--2.29\times {10}^{-3}\text{ Wb}}{0.16\text{ s}}\\ =41.75\times {10}^{-3}\text{ V}\\ \text{=}41.75\times {10}^{-3}\text{ V}\left(\frac{1\text{ mV}}{1.00\text{ V}}\right)\\ =41.75\text{ mV}\end{array}$