A straight wire in a magnetic field experiences a force of 0.017 N when the current in the wire is 1.1 A. The current in the wire is changed, and the wire experiences a force of 0.077 N as a result. What is the new current

The new current in the straight wire is 4.98 A

Explanation:

Given;

initial magnetic force on the wire, F₁ = 0.017 N

initial current flowing on the straight wire, I₁ = 1.1 A

When the current in the wire is changed,

new magnetic force on the wire, F₂ =  0.077 N

the new current in the wire, I₂ = ?

Applying equation of magnetic force on conductor;

F₁ = I₁BLsinθ

F₂ = I₂BLsinθ

BLsinθ  = F₁/I₁ = F₂/I₂

I₂ = (F₂I₁)/F₁

I₂ = (0.077 x 1.1) / 0.017

I₂ = 4.98 A

Therefore, the new current in the straight wire is 4.98 A

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

i don't know that i agree with that but what's the question ?