Chapter 20, Problem 94P

(a)

To determine

The terminal velocity of the rod.

Answer to Problem 94P

The terminal velocity of the rod is $\underset{\_}{3.44\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for motional emf.

    $\epsilon =vBL$

Here, $\epsilon$ is the motional emf, $B$ is the magnetic field, $v$ is the velocity of the rod and $L$ is the length of the rod.

Write the expression to find the magnitude of the magnetic force.

    $F_B=IBL$

Here, $I$ is the current.

Substitute $\frac{\epsilon }{R}$ for $I$ in the above equation.

    $F_B=\left(\frac{\epsilon }{R}\right)BL$

Substitute $vBL$ for $\epsilon$ in the above equation.

    $\begin{array}{c}{F}_B=\left(\frac{vBL}{R}\right)BL\\ =\frac{v{B}^2{L}^2}{R}\end{array}$

Substitute $mg$ for $F_B$ in the above equation and rearrange it to find the velocity $v$.

    $\begin{array}{c}mg=\frac{v{B}^2{L}^2}{R}\\ v=\frac{mgR}{L^2{B}^2}\end{array}$

Here, $m$ is the mass of the rod and $g$ is the acceleration due to gravity.

Conclusion:

Substitute $0.0150\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $8.00\Omega$ for $R$, $1.30\text{m}$ for $L$ and $0.450\text{T}$ for $B$ in the above equation.

    $\begin{array}{c}v=\frac{\left(0.0150\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(8.00\Omega \right)}{{\left(1.30\text{m}\right)}^2{\left(0.450\text{T}\right)}^2}\\ =3.44\text{m}/\text{s}\end{array}$

Therefore, the terminal velocity of the rod is $\underset{\_}{3.44\text{m}/\text{s}}$.

(b)

To determine

The comparison of the rate of gravitational potential energy and the power dissipated in the resistor.

Answer to Problem 94P

The rate of gravitational potential energy and the power dissipated in the resistor are the same, $\underset{\_}{0.505\text{W}}$.

Explanation of Solution

Write the expression for the rate of change of gravitational energy.

    $\frac{\Delta U}{\Delta t}=\frac{mg\Delta y}{\Delta t}$

Here, $\Delta U$ is the change in potential energy.

Substitute $v$ for $\frac{\Delta y}{\Delta t}$ in the above equation.

    $\frac{\Delta U}{\Delta t}=-mgv$

Substitute $\frac{mgR}{L^2{B}^2}$ for $v$ in the above equation.

    $\begin{array}{c}\frac{\Delta U}{\Delta t}=-mg\left(\frac{mgR}{L^2{B}^2}\right)\\ =-\frac{m^2{g}^{2}R}{L^2{B}^2}\end{array}$                                                                                                   (I)

Write the expression for the power converted in the resistor.

    $P=\frac{{\epsilon }^2}{R}$

Substitute $vBL$ for $\epsilon$ in the above equation.

    $P=\frac{{\left(vBL\right)}^2}{R}$

Substitute $\frac{mgR}{L^2{B}^2}$ for $v$ in the above equation.

    $\begin{array}{c}P=\frac{{\left(\frac{mgR}{L^2{B}^2}BL\right)}^2}{R}\\ =\frac{m^2{g}^{2}R}{L^2{B}^2}\end{array}$                                                                                                     (II)

Conclusion:

Substitute $0.0150\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $8.00\Omega$ for $R$, $1.30\text{m}$ for $L$ and $0.450\text{T}$ for $B$ in the equation (I).

    $\begin{array}{c}\frac{\Delta U}{\Delta t}=-\frac{{\left(0.0150\text{kg}\right)}^2{\left(9.80\text{m}/{\text{s}}^2\right)}^2\left(8.00\Omega \right)}{{\left(1.30\text{m}\right)}^2{\left(0.450\text{T}\right)}^2}\\ =-0.505\text{W}\end{array}$

Substitute $0.0150\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^2$ for $g$, $8.00\Omega$ for $R$, $1.30\text{m}$ for $L$ and $0.450\text{T}$ for $B$ in the equation (II).

    $\begin{array}{c}P=\frac{{\left(0.0150\text{kg}\right)}^2{\left(9.80\text{m}/{\text{s}}^2\right)}^2\left(8.00\Omega \right)}{{\left(1.30\text{m}\right)}^2{\left(0.450\text{T}\right)}^2}\\ =0.505\text{W}\end{array}$

Therefore, the rate of gravitational potential energy and the power dissipated in the resistor are the same, $\underset{\_}{0.505\text{W}}$.