Length of Wire Effect on Current and Inner Electric Field.

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SUMMARY

The discussion centers on the effect of wire length on current and electric field strength in two wires connected to a 9-V ideal battery. Wire 2, which is twice the length of wire 1, has double the resistance, leading to half the current compared to wire 1. The electric field strength in wire 2 is also half that of wire 1 due to the same resistivity and cross-sectional area. The solutions manual incorrectly states that the current and electric field are equal in both wires.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with Kirchhoff's Law
  • Knowledge of resistivity and its impact on resistance
  • Basic concepts of electric fields and current density (J = σE)
NEXT STEPS
  • Study the relationship between resistance and wire length in electrical circuits
  • Learn about electric field calculations in conductive materials
  • Explore the implications of resistivity on current flow in different materials
  • Investigate the mathematical principles behind Kirchhoff's Laws
USEFUL FOR

Electrical engineering students, physics enthusiasts, and professionals involved in circuit design and analysis will benefit from this discussion.

alingy1
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>Two wires of the same material are both connected to a 9-V ideal battery. They have the same cross section area but wire 2 is twice longer than wire 1. Compare the following quantities by answering if the quantity for wire 1 is smaller, equal or greater than the same quantity in wire 2.

> - Current in wire 1 vs. current in wire 2.

> - Electric field in wire 1 vs. electric field in wire 2.

The solutions say that current 1 = current 2 and that electric field 1 = electric field 2.

How does the length of the wire affect these two quantities? According to Kirchhoff's Law: $$\Delta V=RI$$.

$$R=\frac{L\rho}{A}\implies$$ Resistance of wire 2 is twice the resistance of wire 1. Therefore, the current in wire 2 will be twice as small as wire 1.

Finally, we know that $$J=\sigma E$$ The resistivity is the same because the material is the same. So, the electric field will be twice as small for wire 2.

The solutions manual seems wrong. Am I right or is there a problem in my reasoning?
 
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Notes.
"twice as small" is ambiguous ... do you mean "half"?
##\Delta V= IR## is Ohm's law.

You can compare values mathematically by dividing them.
 
Last edited:
Hi Simon,
Maybe I was not clear. I was asking if my reasoning, starting by "How does the length" is right. "twice as small"="half" (that is what I meant).
Thanks in advance!
 
... I understood you fine. I was trying to suggest how you may check your own reasoning instead of having to rely on an outside authority.
I could say you are right, but how do you know I'm right?
Instead of relying on other fallable humans, try using maths more formally.
 
alingy1 said:
>Two wires of the same material are both connected to a 9-V ideal battery. They have the same cross section area but wire 2 is twice longer than wire 1. Compare the following quantities by answering if the quantity for wire 1 is smaller, equal or greater than the same quantity in wire 2.

> - Current in wire 1 vs. current in wire 2.

> - Electric field in wire 1 vs. electric field in wire 2.

The solutions say that current 1 = current 2 and that electric field 1 = electric field 2.

How does the length of the wire affect these two quantities? According to Kirchhoff's Law: $$\Delta V=RI$$.

$$R=\frac{L\rho}{A}\implies$$ Resistance of wire 2 is twice the resistance of wire 1. Therefore, the current in wire 2 will be twice as small as wire 1.

Finally, we know that $$J=\sigma E$$ The resistivity is the same because the material is the same. So, the electric field will be twice as small for wire 2.

The solutions manual seems wrong. Am I right or is there a problem in my reasoning?
I see you are from Italy, so English likely is not your first language.
Rather than saying "twice as small", it's better to say something like "half as large", or perhaps say current in wire 1 is twice as large as current in wire 2.

I agree that the solutions manual is wrong.

I would arrive at electric field by considering V = E⋅d .
 

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