# Multistep problem (help on how to do, do not give answers)

• phys-lexic
In summary, we have an 80 m Copper wire with a diameter of 1mm and a 49.0 m iron wire also with a diameter of 1mm joined end to end. With a current of 2A in each wire, we can calculate the current density, electric field, potential difference, and equivalent resistance. The current density is 0.785 A/mm^2 for copper and 0.661 A/mm^2 for iron. The electric field is 37.3 V/m for copper and 9.7 V/m for iron. The potential difference is 3,000 V for copper and 483 V for iron. The equivalent resistance is 1,500 Ω for the combination of copper and
phys-lexic

## Homework Statement

A 80 m Copper wire 1mm in diameter is joined end to end with a 49.0 m iron wire of the same diameter. The current in each is 2 A. (a) Find the current density in each wire (b) Find the electric field in each wire (c) Find the potential difference in each wire (d) Find the equivalent resistance which would carry the same 2A at a potential difference equal to the sum of that across the two. In the answer box,first write values for Cu then a comma(,) and then value for Fe

## The Attempt at a Solution

This is a practice assignment to prep us for our test in weeks, i have an idea how to do these types of problems but this one adds a bunch of mini steps (like for part a, i know that current density is current divided by area, but he's never given a 3d example; we just took the area of a circle, etc., so would i use the entire surface area of the cylinder or just the circle base) and that's where i get confused. could someone, not give me the answers to each step (because i can do bio/chem simply, but i got to learn physics... its just too abstract), but help me out on how to do them to where i can work them out and get the feel of how to approach these kinds.

thank you in advance(a) Current density = 2A/ (πd^2/4) = 0.785 A/mm^2 (Cu), 0.661 A/mm^2 (Fe)(b) Electric field = Current density/resistivity = 37.3 V/m (Cu), 9.7 V/m (Fe)(c) Potential difference = Electric field x Length of wire = 3,000 V (Cu), 483 V (Fe)(d) Equivalent resistance = Potential difference/Current = 1,500 Ω (Cu + Fe)

## 1. What is a multistep problem?

A multistep problem is a complex problem that requires multiple steps or operations to find the solution. It often involves breaking down the problem into smaller parts and using various strategies to solve each part before putting them together to find the final answer.

## 2. How do I approach a multistep problem?

The best way to approach a multistep problem is to read the problem carefully and understand what it is asking. Then, break down the problem into smaller, more manageable parts. Use the given information and any relevant formulas or concepts to solve each part. Finally, check your work and make sure all the steps are correct before arriving at the final answer.

## 3. What are some strategies for solving multistep problems?

Some common strategies for solving multistep problems include working backwards, drawing diagrams or visual representations, making a table or chart to organize information, and using guess and check or trial and error. It is also helpful to identify any patterns or relationships in the problem and use them to your advantage.

## 4. What should I do if I get stuck on a multistep problem?

If you get stuck on a multistep problem, take a step back and review what you have already done. Make sure you understand the given information and the steps you have taken so far. You can also try approaching the problem from a different angle or breaking it down into smaller parts. If you are still struggling, don't be afraid to ask for help from a teacher or classmate.

## 5. How do I check my work on a multistep problem?

To check your work on a multistep problem, make sure you have followed all the steps correctly and have arrived at the correct final answer. You can also plug in your answer to the original problem to see if it makes sense. It is always a good idea to double-check your work and make sure you have not made any careless mistakes.

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