Question

The chocolate crumb mystery.This story begins with Problem 60 in Chapter 23 and continues through Chapters 24 and 25.The chocolate crumb powder moved to the silo through a pipe of radius Rwith uniform speed vand uniform charge density r. (a) Find an expression for the current i(the rate at which charge on the powder moved) through a perpendicular cross section of the pipe. (b) Evaluate ifor the conditions at the factory: pipe radius R = 5.0 cm, speed v = 2.0 m/s ,and charge density p =$\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{C}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$. If the powder were to flow through a change Vin electric potential, its energy could be transferred to a spark at the rate P = iV. (c) Could there be such a transfer within the pipe due to the radial potential difference discussed in Problem 70 of Chapter 24? As the powder flowed from the pipe into the silo, the electric potential of the powder changed. The magnitude of that change was at least equal to the radial potential difference within the pipe (as evaluated in Problem 70 of Chapter 24). (d) Assuming that value for the potential difference and using the current found in (b) above, find the rate at which energy could have been transferred from the powder to a spark as the powder exited the pipe. (e) If a spark did occur at the exit and lasted for 0.20 s (a reasonable expectation), how much energy would have been transferred to the spark? Recall from Problem 60 in Chapter 23 that a minimum energy transfers ofis needed to cause an explosion. (f) Where did the powder explosion most likely occur: in the powder cloud at the unloading bin (Problem 60 of Chapter 25), within the pipe, or at the exit of the pipe into the silo?

Short answer

1. The expression for the current is,$i=p\pi R^{2}v$.
2. The value of the current for the conditions in the factory is$1.7\times {10}^{-5}\mathrm{A}$.
3. There could not be any transfer within the pipe due to the radial potential difference.
4. The rate at which energy could have been transferred from the powder to a spark as the powder exited the pipe is 1.3 W.
5. The energy transferred to the spark is 0.27 J .
6. The spark was likely to have occurred at the exit of the pipe, going into the silo.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• Radius of the pipe is,R = 5 cm = 0.05 m .
• Speed of the pipe is, v = 2 m/s .
• The charge density of the pipe,$\mathrm{p}=1.1\times {10}^{-3}\mathrm{C}/{\mathrm{m}}^3$.
• Time till which the spark lasts, t = 0.20 s.
• Minimum energy transfer required for the explosion, E = 0.15 J.

Understanding the concept of the flow of current and its density

We can find the current using the equation for current in terms of charge density, velocity, and cross-section area. The power due to a force is the rate at which that force does work on an object i.e. the power is the dot product of force and velocity. From this, we can conclude which direction the particle will flow. Power is also the rate of electrical energy transfer. We can write it in terms of potential and current to find its value. From power, we can also calculate the energy at the exit of the pipe.

Formulae:

The current flowing through the pipe is,

i = pAv … (i)

The electric power generated due to the potential difference is,

P = iV … (ii)

The power value relating to speed and force is,

P = F.v … (iii)

The rate of the energy transfer is,

$P=\frac{W}{t}$ … (iv)

The work done by the system due to applied force is,

W = q V … (v)

The work done is,

W = Pt … (vi)

a) Calculation for the expression of the current

From equation (i), we can say that the current is the product of mass density, cross-sectional area, and velocity of the particle. Now, the cross-sectional area of the pipe is circular. Thus,

$A=\pi R^2$

Now, the current flowing through the circuit is given using equation (i) as follows:

$i=p\pi R^{2}v$

Hence, the expression of the current value is $p\pi R^{2}v$.

b) Calculation of the value of the current

By substituting the given values in equation (a), we can get the value of the evaluated current as follows:

$$\begin{gathered} \mathrm{i}=1.1\times {10}^{-3}\mathrm{C}/{\mathrm{m}}^3\times 3.14\times {\left(0.05\mathrm{m}\right)}^2\times 2\mathrm{m}/\mathrm{s} \\ =1.7\times {10}^{-5}\mathrm{A} \end{gathered}$$

Hence, the value of the current is $1.7\times {10}^{-5}\mathrm{A}$.

c) Calculation to check whether there would be any energy transfer

From equation (iii), we can say that the power is the scalar product of force and velocity. Again, we can say that the motion of charge is not in the same direction of potential difference.

Now from equation (ii), we can say that if the potential is zero, the power is also zero.

Thus, the equation (iii) becomes:

$$\begin{gathered} F.v=Fv\cos \theta \\ =0 \end{gathered}$$

It means that F and v are perpendicular to each other. These equations suggest that a radial potential difference and the respective axial flow of particular charge will not together generate/ produce the required transfer of energy to produce/ create a spark.

d) Calculation of the rate of energy transfer from powder to spark

The rate of the energy transfer from the powder form to the spark can be calculated using equation (v) in equation (iv) as follows:

$$\begin{gathered} P=\frac{qV}{t} \\ =iV \end{gathered}$$

Here, the current can be equal to $\frac{q}{t}$. Here assuming the value of potential difference as .

$7.8\times {10}^4\mathrm{V}$

So, we calculate the total power for the rate of the energy transfer using the given values as follows:

$$\begin{gathered} \mathrm{P}=1.7\times {10}^{-5}\mathrm{A}\times 7.8\times {10}^4\mathrm{V} \\ =1.3\mathrm{W} \end{gathered}$$

Hence, the value of the rate of energy transferred to the spark is 1.3 W.

e) Calculation of the energy transferred

So, energy that can be transferred at the exit of the pipe to the spark is given using equation (vi) as follows:

$$\begin{gathered} \mathrm{W}=1.3\mathrm{W}\times 0.20\mathrm{s} \\ =0.27\mathrm{J} \end{gathered}$$

Thus, the energy that can be transferred at the exit of the pipe is 0.27 J.

f) Calculation of the location at which the spark is occurred within the pipe

From condition the energy needed for explosion is0.15 J, but we got energy0.27 J. This result is greater than the0.15 J needed for a spark, so we conclude that the spark was likely to have occurred at the exit of the pipe, going into the silo.