Simple drift speed probem driving me nuts

In summary, the conversation is about solving for the drift speed and number density of electrons in a silver wire with a current of 200 mA and a radius of 0.8mm. The correct answer is 1.07 x 10^-5, but the person is getting 1.06 x 10^-4 and asks for help. It is discovered that they used the wrong value for the density of silver in their calculation.
  • #1
ttran1117
7
0

Homework Statement


A current of 200 mA flows in a silver wire of radius 0.8mm. Find (a) the drift speed of the electrons. (b) the number density of particles.

Homework Equations


The Attempt at a Solution


The answer in the book is 1.07 x 10^-5, but I'm getting 1.06 x 10^-4. Can anyone tell me what I'm doing wrong?

So, I have J = I/A and J=nev_d ==> v_d = J/(ne)
J = 200 x 10^-3/pi(0.8x10^-3)^2

n = p*Avg/M
= 1.05 * 6.022 x 10^23 / 107.868 x 10^-3

plugging in for v_d, I get 1.06 x 10^-4.

This problem is supposed to be simple, but it's driving me nuts! Someone help please! Thanks a lot
 
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  • #2
You used wrong value for the density of silver.

ehild
 
  • #3

It looks like the issue may be with your calculation for the number density of particles. The equation you used, n = p*Avg/M, is for calculating the number density of a gas, not for a solid material like a silver wire. Instead, you should use the equation n = N/V, where N is the number of particles and V is the volume of the wire.

To find N, we can use the formula N = nA, where n is the number of electrons per unit volume (the atomic density) and A is the cross-sectional area of the wire. Since we are dealing with a solid material, we can assume that the number of electrons per atom is equal to the number of protons, so n = Zp, where Z is the atomic number and p is the number of protons.

Plugging in the values, we get N = (47 x 6.022 x 10^23) x (pi(0.8x10^-3)^2) = 1.07 x 10^20.

Now, we can plug this value for N into the equation n = N/V and solve for n. V is equal to the volume of the wire, which can be calculated using the formula V = pi(0.8x10^-3)^2 x L, where L is the length of the wire. Let's assume a length of 1 meter for simplicity, so V = pi(0.8x10^-3)^2 x 1 = 5.03 x 10^-7 m^3.

Plugging in the values, we get n = (1.07 x 10^20)/(5.03 x 10^-7) = 2.13 x 10^26 m^-3. This is the correct value for the number density of particles, which when used in the equation v_d = J/(ne), gives us the correct answer of 1.07 x 10^-5 m/s for the drift speed.

I hope this helps clear up any confusion and solves the problem for you. If you have any further questions, please don't hesitate to ask. Keep up the good work!
 

Related to Simple drift speed probem driving me nuts

What is a simple drift speed problem?

A simple drift speed problem involves calculating the average speed of particles in a material when an electric field is applied. This can be used to understand the movement of charged particles in a circuit or other conductive material.

Why is this problem important to study?

Understanding drift speed can help us understand how electricity flows through conductive materials, which is essential for many technological applications such as electronics and power transmission.

What factors affect drift speed?

The drift speed of particles is affected by the strength of the electric field, the density and type of particles in the material, and the temperature of the material.

How do you calculate drift speed?

Drift speed can be calculated by dividing the current (I) by the product of the material's cross-sectional area (A) and the number of particles per unit volume (n). This is represented by the equation v = I/(An).

What are some real-world applications of drift speed?

Drift speed is used in various technologies, including electronics, power transmission, and particle accelerators. It also plays a role in understanding the movement of charged particles in the Earth's atmosphere and in space.

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