Chapter 23, Problem 56P

(a)

To determine

The expression for the current in the light bulb

Answer to Problem 56P

The expression for the current is $I=Blv/R$.

Explanation of Solution

Write the equation for the current in the bulb in term of the emf.

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

Here, $\epsilon$ is the emf in the bulb and $R$ is the resistance of the bulb. Write the equation for the emf of the bulb in terms of magnetic field.

    $\epsilon =Blv$        (II)

Here, $B$ is the magnetic field in which the system is immersed, $l$ is the distance between the rails and $v$ is the speed of movement of the bar.

Conclusion:

Substitute equation (II) in equation (I).

    $I=\frac{Blv}{R}$        (III)

Therefore, the expression for the current in the light bulb as a function of $B$, $l$, $R$ and $v$ is $I=Blv/R$.

(b)

To determine

The analysis model describing the moving bar

Answer to Problem 56P

The analysis model describing the moving when the bulb has maximum power is particle under equilibrium.

Explanation of Solution

Write the equation for the power of the light bulb.

    $P=Fv$

Here, $F$ is the force on the moving bar and $v$ is the speed of the moving bar.

From the above equation, both the force and the velocity of the moving have to be maximum for the power on the light bulb to be maximum. The condition of maximum power points to energy loss which could happen only for a particle in equilibrium.

Conclusion:

Therefore, the analysis model describing the moving when the bulb has maximum power is particle under equilibrium.

(c)

To determine

The speed of the bar

Answer to Problem 56P

The speed of the bar at maximum power is $281\text{m/s}$.

Explanation of Solution

The magnetic flux points into the page thereby making the counterclockwise current to move out of the page. Write the equation for the magnetic force that the current is flowing upwards in the bar.

    $F=IlB$        (IV)

Here, $I$ is the current flowing upwards, $l$ is distance between the two rails and $B$ is the magnetic field. Substitute equation (III) in equation (IV) and rearrange to find the equation for the speed of the bar.

    $\begin{array}{c}F=I\left(\frac{Blv}{R}\right)B\\ F=\frac{B^2{l}^2}{R}v\\ v=\frac{FR}{B^2{l}^2}\end{array}$        (V)

Conclusion:

Substitute $0.600\text{N}$ for $F_B$, $48.0\Omega$ for $R$, $0.400\text{T}$ for $B$ and $0.800\text{m}$ for $l$.

    $\begin{array}{c}v=\frac{\left(0.600\text{N}\right)\left(48.0\Omega \right)}{{\left(0.400\text{T}\right)}^2{\left(0.800\text{m}\right)}^2}\\ =281\text{m/s}\end{array}$

Therefore, the speed of the bar at maximum power is $281\text{m/s}$.

(d)

To determine

The current in the bulb

Answer to Problem 56P

The current in the light bulb at maximum power is $1.88\text{A}$.

Explanation of Solution

Substitute equation (V) in equation (III).

    $\begin{array}{c}I=\frac{Bl}{R}\left(\frac{FR}{B^2{l}^2}\right)\\ =\frac{F}{Bl}\end{array}$        (VI)

Conclusion:

Substitute $0.600\text{N}$ for $F$, $0.400\text{T}$ for $B$ and $0.800\text{m}$ for $l$.

    $\begin{array}{c}I=\frac{0.600\text{N}}{\left(0.400\text{T}\right)\left(0.800\text{m}\right)}\\ =1.88\text{A}\end{array}$

Therefore, the current in the light bulb at maximum power is $1.88\text{A}$.

(e)

To determine

The maximum power delivered

Answer to Problem 56P

The maximum power delivered to the light bulb is $1.69\text{W}$.

Explanation of Solution

Write the equation for the power in the light bulb.

    $P=I^{2}R$        (VII)

Here, $I$ is the current flow to the bulb and $R$ is the resistance of the bulb. Substitute equation (VI) in equation (VII).

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

Conclusion:

Substitute $0.600\text{N}$ for $F$, $0.400\text{T}$ for $B$, $0.800\text{m}$ for $l$ and $48.0\Omega$ for $R$.

    $\begin{array}{c}P={\left(\frac{0.600\text{N}}{\left(0.400\text{T}\right)\left(0.800\text{m}\right)}\right)}^2\left(48.0\Omega \right)\\ =\left(3.52\times 48.0\right)\text{W}\\ =\text{169}\text{W}\end{array}$

Therefore, the maximum power delivered to the light bulb is $1.69\text{W}$.

(f)

To determine

The maximum input power delivered

Answer to Problem 56P

The maximum input power delivered to the bar is $1.69\text{W}$.

Explanation of Solution

Write the equation for the power delivered to the bar.

    $P=Fv$        (VIII)

Here, $F$ is the force on the bar and $v$ is the speed of the bar. Substitute equation (V) in equation (VIII).

    $P=\frac{F^{2}R}{B^2{l}^2}$

Conclusion:

Substitute $0.600\text{N}$ for $F$, $48.0\Omega$ for $R$, $0.400\text{T}$ for $B$ and $0.800\text{m}$ for $l$.

    $\begin{array}{c}P=\frac{{\left(0.600\text{N}\right)}^2\left(48.0\Omega \right)}{{\left(0.400\text{T}\right)}^2{\left(0.800\text{m}\right)}^2}\\ =169\text{W}\end{array}$

Therefore, the maximum input power delivered to the bar is $1.69\text{W}$.

(g)

To determine

Whether the speed changes or not

Answer to Problem 56P

The speed changes when the resistance increases.

Explanation of Solution

Write the equation for the speed of the bar from equation (V).

    $\begin{array}{c}v=\frac{FR}{B^2{l}^2}\\ v\propto R\end{array}$

Hence, the speed of the bar and the resistance are proportional to each other, given, all other quantities are kept constant.

Conclusion:

Therefore, the speed of the bar changes when the resistance increases.

(h)

To determine

Whether the speed increases or decreases

Answer to Problem 56P

The speed changes increases the resistance increases.

Explanation of Solution

Write the equation for the speed of the bar from equation (V).

    $\begin{array}{c}v=\frac{FR}{B^2{l}^2}\\ v\propto R\end{array}$

Hence, the speed of the bar and the resistance are proportional to each other, given, all other quantities are kept constant.

Conclusion:

Therefore, the speed of the bar increases when the resistance increases.

(i)

To determine

Whether the power changes

Answer to Problem 56P

The power changes when the current increases

Explanation of Solution

An increase in current leads to a change in the mechanical load as the current as the current is analogous to mechanical load.

The mechanical power depends on the load. Therefore, the change in current will lead to change in the power.

Conclusion:

Therefore, the power changes when the current increases.

(j)

To determine

Whether the power is smaller or larger

Answer to Problem 56P

The power changes and becomes larger.

Explanation of Solution

According to ohm’s law, the current and resistance are inversely proportional to each other. If the current and resistance has to increase together, the load has to increase further.

The increase in the load will increase the power.

Conclusion:

Therefore, the power changes and becomes larger when the current and resistance increases.