# Electric field in the narrow wire

• DottZakapa
In summary, the conversation discusses the use of Ohm's law and the formula for resistance in finding the ratio of potential differences in two wires with different current densities and resistances. The strategy is to write Ohm's law for each wire separately and then find the ratio by using the formula for resistance. The conversation ends with the realization that considering the resistance of the wires is necessary in order to find the correct ratio.
DottZakapa
Homework Statement
A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is
Relevant Equations
current
i've started from this I1=I2
then
I1= JA1=##\frac {E l} R##

I2= JA2=##\frac {E_2 l} R##

but can't get anything useful relating them. Am i forgetting any other useful formula?
I get as result E4

Why are you using the same J in the two equations? If the current is the same in the two wires would the current density also be the same? Also the wires don't have the same resistance.

kuruman said:
Why are you using the same J in the two equations? If the current is the same in the two wires would the current density also be the same? Also the wires don't have the same resistance.
you right
##I_1= J A## , ##A_2 = pi 4b^2##
##I_2=J_2 A_2##
##j_2= \frac j 4 ##
then ?
##I_2 = \frac j 4 \ A_2 = \frac {E l} {R}\space but\space then \space A_2\space simplify\space with\space the\space 4\space and \space i\space get\space back\space to\space the\space beginning...##
sorry I'm not getting anywhere

There are many equivalent ways to get to the answer, but you need a goal and a strategy. The goal is to find the ratio ##E_2/E##. The strategy is the use of Ohm's law and the formula for the resistance of a cylindrical wire ##R=\dfrac{\rho L}{A}##. You need to write Ohm's law for each wire separately then write separate expressions for each wire's resistance and then take the appropriate ratio.

DottZakapa
kuruman said:
There are many equivalent ways to get to the answer, but you need a goal and a strategy. The goal is to find the ratio ##E_2/E##. The strategy is the use of Ohm's law and the formula for the resistance of a cylindrical wire ##R=\dfrac{\rho L}{A}##. You need to write Ohm's law for each wire separately then write separate expressions for each wire's resistance and then take the appropriate ratio.
now it works, thanks a lot for your time

kuruman

## 1. What is an electric field in a narrow wire?

The electric field in a narrow wire refers to the distribution of electric charges along the length of a thin wire. It is a vector quantity that describes the strength and direction of the electric force that a charged particle would experience if placed at a particular point along the wire.

## 2. How is the electric field in a narrow wire calculated?

The electric field in a narrow wire can be calculated using the formula E = (λ/2πε₀r) where E is the electric field, λ is the linear charge density of the wire, ε₀ is the permittivity of free space, and r is the distance from the wire.

## 3. What is the relationship between the electric field and the charge density in a narrow wire?

The electric field in a narrow wire is directly proportional to the linear charge density of the wire. This means that as the charge density increases, the electric field also increases.

## 4. How does the electric field in a narrow wire affect nearby charged particles?

The electric field in a narrow wire can exert a force on nearby charged particles, causing them to either repel or attract depending on their charge. This force is strongest when the charged particles are closer to the wire and decreases as they move further away.

## 5. How does the electric field in a narrow wire differ from that of a wider wire?

The electric field in a narrow wire is stronger than that of a wider wire, as the charges are more concentrated in a smaller area. Additionally, the electric field in a narrow wire is more uniform along the length of the wire, whereas a wider wire may have variations in the electric field due to its larger surface area.

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