Chapter 32, Problem 79PQ

A conducting single-turn circular loop with a total resistance of 5.00 Ω is placed in a time-varying magnetic field that produces a magnetic flux through the loop given by Φ_B = a + bt² – ct³, where a = 4.00 Wb, b = 11.0 Wb/s^–2, and c = 6.00 Wb/s^–3. Φ_B is in webers, and t is in seconds. What is the maximum current induced in the loop during the time interval t = 0 to t = 3.50 s?

To determine

The maximum current induced in the loop.

Answer to Problem 79PQ

The maximum induced current is $1.34\text{A}$.

Explanation of Solution

Write the expression for the induced emf.

    $\epsilon =\frac{d{\varphi }_{\text{B}}}{dt}$                                                                                                           (I)

Here, $\epsilon$ is the induced emf and ${\varphi }_{\text{B}}$ is the magnetic flux.

Write the relation between the induced emf and the induced current.

    $I=\frac{\epsilon }{R}$                                                                                                               (II)

Here, $I$ is the induced current, $\epsilon$ is the induced emf and $R$ is the resistance.

Write the expression for the magnetic flux through the loop.

    ${\varphi }_{\text{B}}=a+b{t}^2+c{t}^3$

Conclusion:

Substitute $4.00\text{Wb}$ for $a$, $11.0\text{Wb}/{\text{s}}^{-2}$ for $b$ and $6.00\text{Wb}/{\text{s}}^{-3}$ for $c$.

    ${\varphi }_{\text{B}}=\left(4.00\text{Wb}\right)+\left(11.0\text{Wb}/{\text{s}}^{-2}\right)t^2+\left(6.00\text{Wb}/{\text{s}}^{-3}\right)t^3$                                    (III)

Substitute equation (III) in the equation (I).

    $\begin{array}{c}\epsilon =\frac{d}{dt}\left[\left(4.00\text{Wb}\right)+\left(11.0\text{Wb}/{\text{s}}^{-2}\right)t^2+\left(6.00\text{Wb}/{\text{s}}^{-3}\right)t^3\right]\\ =\left[2\times \left(11.0\text{Wb}/{\text{s}}^{-2}\right)t-3\times \left(6.00\text{Wb}/{\text{s}}^{-3}\right)t^2\right]\end{array}$                            (IV)

Differentiate the equation (IV) with respect to time.

    $\frac{d\epsilon }{dt}=\frac{d}{dt}\left[2\times \left(11.0\text{Wb}/{\text{s}}^{-2}\right)t-3\times \left(6.00\text{Wb}/{\text{s}}^{-3}\right)t^2\right]$

Substitute $0$ for $\frac{d\epsilon }{dt}$ in the above equation.

    $\begin{array}{c}0=\frac{d}{dt}\left[2\times \left(11.0\text{Wb}/{\text{s}}^{-2}\right)t-3\times \left(6.00\text{Wb}/{\text{s}}^{-3}\right)t^2\right]\\ 0=2\times \left(11.0\text{Wb}/{\text{s}}^{-2}\right)-6\times \left(6.00\text{Wb}/{\text{s}}^{-3}\right)t\\ t=\frac{2\times \left(11.0\text{Wb}/{\text{s}}^{-2}\right)}{6\times \left(6.00\text{Wb}/{\text{s}}^{-3}\right)}\\ =0.611\text{s}\end{array}$

Substitute $0.611\text{s}$ for $t$ in the equation-(IV) to find $\epsilon$.

    $\begin{array}{c}\epsilon =\left[2\times \left(11.0\text{Wb}/{\text{s}}^{-2}\right)\times 0.611\text{s}-3\times \left(6.00\text{Wb}/{\text{s}}^{-3}\right)\times {\left(0.611\text{s}\right)}^2\right]\\ =6.722\text{V}\end{array}$

Substitute $6.722\text{V}$ for $\epsilon$ and $5.00\Omega$ for $R$ in the equation-(II) to find $I$.

    $\begin{array}{c}I=\frac{6.722\text{V}}{5.00\Omega }\\ =1.34\text{A}\end{array}$

Therefore, the maximum induced current is $1.34\text{A}$.