Chapter 21, Problem 10P

A circular loop in the plane of the paper lies in a 0.65-T magnetic field pointing into the paper. The loop's diameter changes from 20.0 cm to 6.0 cm in 0.50 s. What is (a) the direction of the induced current, (6) the magnitude of the average induced emf_,and (c) the average induced current if the coil resistance is 2.5 O?

To determine

Part a:

The direction of the induced current

Answer to Problem 10P

Solution:

The direction of the induced current is clockwise.

Explanation of Solution

Given:

A circular loop in the plane of paper

The magnetic field is 0.65 T pointing into the paper.

The resistance of the coil is $2.5\Omega$.

The diameter changes from 2.0 cm to 6.0 cm in 0.50 s.

According to the Lenz’s law, the direction of the induced current is such that it opposes the change in the magnetic field that produces it.

According to the right-hand grab rule, the magnetic field will point into the plane of paper, only, when the direction of the current is clockwise.

To determine

Part b:

The magnitude of the average induced EMF.

Answer to Problem 10P

Solution:

The magnitude of the average induced EMF is equal to $3.3\text{ mV}$.

Explanation of Solution

Given:

A circular loop in the plane of paper

The magnetic field is 0.65 T pointing into the paper.

The resistance of the coil is $2.5\Omega$.

The diameter changes from 2.0 cm to 6.0 cm in 0.50 s.

Formula used:

The formula to calculate the flux through the circular loop is as follows:

$\varphi =BA\cos \theta$

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

The area of the circular loop is $\pi r^2$, that 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, r is the radius and d is the diameter.

Therefore, the expression for the flux is rewritten as follows:

$\begin{array}{c}\varphi =B\pi {\left(\frac{d}{2}\right)}^2\cos \theta \\ =\frac{B\pi d^2\cos \theta }{4}\end{array}$

Therefore,

$\varphi =\frac{B\pi d^2\cos \theta }{4}$

The expression for the induced EMF is as follows:

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

Here, $d\varphi$ is the change in the flux, N is the number of turns, and $dt$ is the change in the time.

Calculation:

Initially calculate the change in the flux as follows:

$d\varphi ={\varphi }_2-{\varphi }_1$

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

Substitute $\frac{B\pi d_{{}^2}^2\cos \theta }{4}$ for ${\varphi }_2$ and $\frac{B\pi d_{{}^1}^2\cos \theta }{4}$ for ${\varphi }_1$ in the equation $d\varphi ={\varphi }_2-{\varphi }_1$.

$\begin{array}{c}d\varphi =\frac{B\pi d_{{}^2}^2\cos \theta }{4}-\frac{B\pi d_{{}^1}^2\cos \theta }{4}\\ =\frac{B\pi \cos \theta }{4}\left(d_{{}^2}^2-d_{{}^1}^2\right)\end{array}$

Therefore, the change in flux is as follows:

$d\varphi =\frac{B\pi \cos \theta }{4}\left(d_{{}^2}^2-d_{{}^1}^2\right)$

Here, $d_1$ is the diameter initially and $d_2$ is the diameter finally.

Substitute 2.0 cm for $d_1$, 6.0 cm for $d_2$, 0.65 T for B, 3.14 for $\pi$, and $0°$ for $\theta$ in the equation $d\varphi =\frac{B\pi \cos \theta }{4}\left(d_{{}^2}^2-d_{{}^1}^2\right)$.

$\begin{array}{c}d\varphi =\frac{\left(0.65\text{ T}\right)\left(3.14\right)}{4}\left[{\left(6.0\text{ cm}\right)}^2-{\left(2.0\text{ cm}\right)}^2\right]\\ =\frac{\left(0.65\text{ T}\right)\left(3.14\right)}{4}\left[{\left(6.0\text{ cm}\right)}^2{\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}^2-{\left(2.0\text{ cm}\right)}^2{\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}^2\right]\\ =\left(0.51\text{ T}\right)\left(0.0036{\text{ m}}^2-0.0004{\text{ m}}^2\right)\\ =0.0016\text{ Wb}\end{array}$

Now, substitute $0.0016\text{ Wb}$ for $d\varphi$, 1 for N, and 0.50 s for $dt$ in the equation $\epsilon =-N\frac{d\varphi }{dt}$.

$\begin{array}{c}\epsilon =-\left(1\right)\frac{0.0016\text{ Wb}}{0.50\text{ s}}\\ =-0.0033\text{ V}\\ =\text{3}\text{.3 mV}\end{array}$

To determine

Part c:

The average induced current.

Answer to Problem 10P

Solution:

The average induced current is $0.0013\text{ A}$.

Explanation of Solution

Given:

A circular loop in the plane of paper

The magnetic field is 0.65 T pointing into the paper.

The resistance of the coil is $2.5\Omega$.

The diameter changes from 2.0 cm to 6.0 cm in 0.50 s.

Formula used:

The expression to calculate the average induced current is as follows:

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

Here, $\epsilon$ is the EMF and R is the resistance of the coil.

Calculation:

Substitute $3.3\text{ mV}$ for $\epsilon$ and $2.5\Omega$ for R in the equation $I=\frac{\epsilon }{R}$.

$\begin{array}{c}I=\frac{3.3\text{ mV}}{2.5\Omega }\\ =1.32\text{ mA}\\ =1.32\text{ mA}\left(\frac{{10}^{-3}\text{ A}}{1.00\text{ mA}}\right)\\ =0.0013\text{ A}\end{array}$