Chapter 31, Problem 58AP

(a)

To determine

The expression for the current in terms of $B$, $l$ , $R$ and $v$.

Answer to Problem 58AP

The expression for the current in terms of $B$, $l$ , $R$ and $v$ is $I=\frac{Blv}{R}$.

Explanation of Solution

Write the expression to obtain the area enclosed between the bar and the two rails.

    $A=lx$

Here, $A$ is the area enclosed between the bar and the two rails, $l$ is the length of the bar and $x$ is the rod displacement.

Write the expression to obtain the magnetic flux.

    ${\varphi }_{\text{B}}=BA$

Here, ${\varphi }_{\text{B}}$ is the magnetic flux, $B$ is the magnetic field and $A$ is the area enclosed between the bar and the two rails.

Substitute $lx$ for $A$ in the above equation.

    ${\varphi }_{\text{B}}=Blx$

Write the expression for induced emf based on Faraday’s law.

    $\epsilon =-\frac{d{\varphi }_{\text{B}}}{dt}$

Here, $\epsilon$ is the induced emf and $\frac{d{\varphi }_{\text{B}}}{dt}$ is the rate of change of flux.

Substitute $Blx$ for ${\varphi }_{\text{B}}$ in the above expression.

    $\begin{array}{c}\epsilon =-\frac{d\left(Blx\right)}{dt}\\ =-Bl\frac{dx}{dt}\\ =-Blv\end{array}$

Here, $v$ is the change in $x$ with time $t$.

Write the expression to obtain the magnitude of current induced in the bar.

    $I=\frac{\left|\epsilon \right|}{R}$

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

Substitute $-Blv$ for $\epsilon$ in the above equation.

    $\begin{array}{c}I=\frac{\left|-Blv\right|}{R}\\ =\frac{Blv}{R}\end{array}$

Therefore, the expression for the current in terms of $B$, $l$, $R$ and $v$ is $I=\frac{Blv}{R}$.

(b)

To determine

The analysis model that properly describes the moving bar when maximum power is delivered to the light bulb.

Answer to Problem 58AP

The bar moves with the constant velocity when maximum power is delivered to the light bulb. This constant velocity with which the bar is moving is due to the reason that the magnetic force is equal in magnitude to the applied force but they are opposite in direction. Thus this analysis model is similar model of particle in equilibrium.

Explanation of Solution

The analysis model that properly describes the moving bar when maximum power is delivered to the light bulb is similar to the model of particle in equilibrium that is when the resultant of all the forces is equal to zero. It means that the particle is in equilibrium.

The bar moves with the constant velocity when maximum power is delivered to the light bulb. This constant velocity with which the bar is moving is due to the reason that the magnetic force is equal in magnitude to the applied force but they are opposite in direction. Thus this analysis model is similar model of particle in equilibrium.

(c)

To determine

The speed with which the bar is moving when maximum power is delivered to the light bulb.

Answer to Problem 58AP

The speed with which the bar is moving when maximum power is delivered to the light bulb is $281.25\text{m}/\text{s}$.

Explanation of Solution

Write the expression when maximum power is delivered to the light bulb.

    $F=F_{\text{B}}$                                                   (I)

Here, $F$ is the applied force and $F_{\text{B}}$ is the magnetic force.

Write the expression to obtain the applied force to the bar.

    $F=BIl$                                               (II)

Here, $F$ is the applied force, $B$ is the magnetic field, $I$ is the current in the light bulb and $l$ is the length of the bar.

Compare equation (I) and (II).

    $F_{\text{B}}=BIl$                                                (III)

Substitute $\frac{Blv}{R}$ for $I$ in the above expression.

    $\begin{array}{c}{F}_{\text{B}}=B\left(\frac{Blv}{R}\right)l\\ F_{\text{B}}=\frac{B^2{l}^{2}v}{R}\\ v=\frac{F_{\text{B}}R}{B^2{l}^2}\end{array}$                                            (IV)

Here, $v$ is the speed with which bar is moving in magnetic field and $R$ is the resistance.

Conclusion:

Substitute $48.0\text{Ω}$ for $R$, $0.400\text{T}$ for $B$, $0.800\text{m}$ for $l$ and $0.600\text{N}$ for $F$ in equation (IV) to calculate $v$.

    $\begin{array}{c}v=\frac{\left(0.600\text{N}\right)\left(48.0\text{Ω}\right)}{{\left(0.400\text{T}\right)}^2{\left(0.800\text{m}\right)}^2}\\ =281.25\text{m}/\text{s}\end{array}$

Therefore, the speed with which the bar is moving when maximum power is delivered to the light bulb is $281.25\text{m}/\text{s}$.

(d)

To determine

The current in the light bulb when maximum power is delivered to it.

Answer to Problem 58AP

The current in the light bulb when maximum power is delivered to it is $1.875\text{A}$.

Explanation of Solution

Consider equation (III).

    $F_{\text{B}}=BIl$

Re-write the above equation.

    $I=\frac{F_{\text{B}}}{Bl}$                                           (V)

Conclusion:

Substitute $0.600\text{N}$ for $F_{\text{B}}$, $0.400\text{T}$ for $B$ and $0.800\text{m}$ for $l$ in equation (V) to calculate $I$.

    $\begin{array}{c}I=\frac{0.600\text{N}}{\left(0.400\text{T}\right)\left(0.800\text{m}\right)}\\ =1.875\text{A}\end{array}$

Therefore, the current in the light bulb when maximum power is delivered to it is $1.875\text{A}$.

(e)

To determine

The maximum power delivered to the light bulb.

Answer to Problem 58AP

The maximum power delivered to the light bulb is $169\text{W}$.

Explanation of Solution

Write the expression to obtain the maximum power delivered to the light bulb.

    $P=I^{2}R$

Here, $P$ is the maximum power delivered to the light bulb, $I$ is the current and $R$ is the resistance.

Conclusion:

Substitute $1.875\text{A}$ for $I$ and $48.0\text{Ω}$ for $R$ in the above equation.

    $\begin{array}{c}P={\left(1.875\text{A}\right)}^2\left(48.0\text{Ω}\right)\\ =168.75\text{W}\\ \approx \text{169}\text{W}\end{array}$

Therefore, the maximum power delivered to the light bulb is $169\text{W}$.

(f)

To determine

The maximum mechanical input power delivered to the bar by the applied force.

Answer to Problem 58AP

The maximum mechanical input power delivered to the bar by the applied force is $169\text{W}$.

Explanation of Solution

Write the expression to obtain the maximum mechanical input power delivered to the bar by the applied force.

    $P_{\text{m}}=Fv$

Here, $P_{\text{m}}$ is the maximum mechanical input power delivered to the bar by the applied force and $F$ is the applied force and $v$ is the speed with which bar is moving.

Conclusion:

Substitute $0.600\text{N}$ for $F$ and $281.25\text{m}/\text{s}$ for $v$ in the above equation to calculate $P_{\text{m}}$.

    $\begin{array}{c}{P}_{\text{m}}=\left(0.600\text{N}\right)\left(281.25\text{m}/\text{s}\right)\\ =168.75\text{W}\\ \approx 169\text{W}\end{array}$

Therefore, the maximum mechanical input power delivered to the bar by the applied force is $169\text{W}$.

(g)

To determine

Weather the speed found in part (c) change, if the resistance increases and all the other quantity remain same.

Answer to Problem 58AP

Yes, the speed found in part (c) change, if the resistance increases and all the other quantity remain same.

Explanation of Solution

Write the expression (IV) used in part (c) to determine the speed of the bar.

    $v=\frac{FR}{B^2{l}^2}$

From the above expression, speed is directly proportional to the resistance. Thus, if the resistance increases than the speed is also increases if all the other quantity remains constant.

Therefore, the speed found in part (c) changes, if the resistance increases and all the other quantity remains the same.

(h)

To determine

Weather the speed found in part (c) increases or decreases if resistance increases and all the other quantities remain same.

Answer to Problem 58AP

The speed found in part (c) increases if resistance increases and all the other quantities remain same.

Explanation of Solution

Write the expression (IV) used in part (c) to determine the speed of the bar.

    $v=\frac{FR}{B^2{l}^2}$

From the above expression, speed is directly proportional to the resistance.

Therefore, the speed found in part (c) increases if resistance increases and all the other quantities remain same.

(i)

To determine

Weather the power found in part (f) change if resistance increases as current increases.

Answer to Problem 58AP

Yes, the power found in part (f) changes if resistance increases as current increases.

Explanation of Solution

Write the expression used in part (f) to determine the speed of the bar.

    $P=Fv$

From the above expression, power is directly proportional to the velocity and further velocity is directly proportional to the resistance.

Therefore, the power found in part (f) changes if resistance increases as current increases.

(j)

To determine

Weather the power found in part (f) is larger or smaller if resistance increases as current increases.

Answer to Problem 58AP

The power found in part (f) is smaller if the resistance increases as current increases.

Explanation of Solution

Write the expression used in part (f) to determine the speed of the bar.

    $P=Fv$

From the above expression, power is directly proportional to the velocity and further velocity is directly proportional to the resistance.

Thus, power increase if resistance increases.

Therefore, the power found in part (f) is smaller if the resistance increases as current increases.