Chapter 32, Problem 78PQ

To determine

The electric power in the circuit at $t=0$.

Answer to Problem 78PQ

The electric power in the circuit at $t=0$ is $\frac{{\left(\pi r^2\right)}^2{B}_0^2}{R}$.

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 expression for the magnetic flux.

    ${\varphi }_{\text{B}}=B\left(t\right)A$                                                                                                     (II)

Here, $B\left(t\right)$ is the time varying magnetic field and $A$ is the area of the loop.

Write the expression for the electric power.

    $P=\frac{{\epsilon }^2}{R}$                                                                                                           (III)

Here, $P$ is the electric power and $R$ is the resistance.

Substitute $B\left(t\right)A$ for ${\varphi }_B$ in the equation (I).

  $\epsilon =-\frac{d\left(B\left(t\right)A\right)}{dt}$                                                                                             (IV)

Substitute $B_0{e}^{-t}$ for $B\left(t\right)$ and $\pi r^2$ for $A$ in the equation (IV).

    $\epsilon =-\frac{d\left(B_0{e}^{-t}\right)}{dt}\left(\pi r^2\right)$

Here, $r$ is the radius of the circular loop, $t$ is the time and $B_0$ is the constant.

Differentiate the above equation with respect to time.

  $\begin{array}{c}\epsilon =-\frac{d\left(B_0{e}^{-t}\right)}{dt}\left(\pi r^2\right)\\ =-\left(B_0{e}^{-t}\right)\left(-1\right)\left(\pi r^2\right)\\ =\left(\pi r^2\right)B_0{e}^{-t}\end{array}$

Conclusion:

Substitute $\left(\pi r^2\right)B_0{e}^{-t}$ for $\epsilon$ in the equation (III).

    $\begin{array}{c}P=\frac{{\left(\left(\pi r^2\right)B_0{e}^{-1}\right)}^2}{R}\\ =\frac{{\left(\pi r^2\right)}^2{B}_0^2{e}^{-2t}}{R}\end{array}$                                                                                       (V)

Substitute $0$ for $t$ in the equation (V) to find $P$.

    $\begin{array}{c}P=\frac{{\left(\pi r^2\right)}^2{B}_0^2{e}^{-2\left(0\right)}}{R}\\ =\frac{{\left(\pi r^2\right)}^2{B}_0^2}{R}\end{array}$

Therefore, the electric power in the circuit at $t=0$ is $\frac{{\left(\pi r^2\right)}^2{B}_0^2}{R}$.