Math Problem Statement

ONAL IN

E

OF PHY

  1. Current happenings (5 points)

Answer the following questions:

A. (3 points) Nerve cells transmit signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions, each with charge +e, into the axon. Measurements have revealed that typically about 5.6 x 1011 Na+ ions enter each meter of the axon during a time of 10 [ms]. What is the current during this inflow of charge in a meter of axon?

B. (2 points) A wire with resistivity 3.54 x 10-7 [m] has a cross-sectional area of 2.33 [mm]. It carries a current of 2.00 [A]. What is the magnitude of the electric field throughout the wire?

U.P.

DIL

MAN e=1.60x10^-19 C

Solution

Let's break this down into two parts:

Part A: Current During Inflow of Na+ Ions

We are given the following information:

  • The number of Na+ ions entering the axon per meter: 5.6×10115.6 \times 10^{11} ions.
  • The time over which they enter: 10 ms = 10×10310 \times 10^{-3} s.
  • The charge of each Na+ ion: +e=1.60×1019+e = 1.60 \times 10^{-19} C.

The current II is defined as the rate of flow of charge. To find it, we first calculate the total charge QQ that enters the axon, and then use the formula for current:

I=QtI = \frac{Q}{t}

where:

  • Q=(Number of ions)×(Charge per ion)Q = \text{(Number of ions)} \times \text{(Charge per ion)}
  • tt is the time in seconds.

Step 1: Calculate the total charge

Q=(5.6×1011ions)×(1.60×1019C/ion)=8.96×108CQ = (5.6 \times 10^{11} \, \text{ions}) \times (1.60 \times 10^{-19} \, \text{C/ion}) = 8.96 \times 10^{-8} \, \text{C}

Step 2: Calculate the current

Now that we know the total charge, we can calculate the current by dividing it by the time:

I=8.96×108C10×103s=8.96×105AI = \frac{8.96 \times 10^{-8} \, \text{C}}{10 \times 10^{-3} \, \text{s}} = 8.96 \times 10^{-5} \, \text{A}

So, the current during the inflow of Na+ ions is 8.96×105A8.96 \times 10^{-5} \, \text{A}.


Part B: Magnitude of the Electric Field in a Wire

We are given:

  • Resistivity of the wire: ρ=3.54×107Ωm\rho = 3.54 \times 10^{-7} \, \Omega \cdot \text{m}.
  • Cross-sectional area of the wire: A=2.33mm2=2.33×106m2A = 2.33 \, \text{mm}^2 = 2.33 \times 10^{-6} \, \text{m}^2.
  • Current through the wire: I=2.00AI = 2.00 \, \text{A}.

To find the electric field EE, we use Ohm’s law in the form:

E=IρAE = \frac{I \cdot \rho}{A}

Step 1: Calculate the electric field

Substitute the given values:

E=(2.00A)×(3.54×107Ωm)2.33×106m2E = \frac{(2.00 \, \text{A}) \times (3.54 \times 10^{-7} \, \Omega \cdot \text{m})}{2.33 \times 10^{-6} \, \text{m}^2}

E=7.08×107V/m2.33×106=0.304V/mE = \frac{7.08 \times 10^{-7} \, \text{V/m}}{2.33 \times 10^{-6}} = 0.304 \, \text{V/m}

Thus, the magnitude of the electric field in the wire is 0.304 V/m.


Summary of Results:

  • Part A: The current during the inflow of Na+ ions is 8.96×105A8.96 \times 10^{-5} \, \text{A}.
  • Part B: The magnitude of the electric field in the wire is 0.304 V/m.

Would you like more details or have further questions on this? Here are a few more related questions for you to explore:

  1. What is the relationship between current and charge in a conductor?
  2. How does resistivity affect the electric field in a material?
  3. What happens to the current if the number of Na+ ions entering the axon changes?
  4. How would the electric field change if the cross-sectional area of the wire was increased?
  5. How does the time duration of the Na+ ion inflow affect the overall current?

Tip: Always remember that current is the rate at which charge flows, and electric field is related to the potential difference and resistivity of materials!

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Math Problem Analysis

Mathematical Concepts

Current
Charge
Electric Field
Resistivity

Formulas

I = Q / t
E = I * ρ / A

Theorems

Ohm's Law
Current-Flow Theorem

Suitable Grade Level

Grades 11-12