Two 3 m long conductors are separated from each other by 5 mm and carry a current of 10 A dc. Calculate the force that one conductor exerts on the other.

0.012 N

Explanation:

The formula to apply is that adopted from the Ampere law which is;

F= μ* I₁*I₂*l /2πd   where

F is force that one conductor exerts on the other.

μ = magnetic permeability of free space = 4π×10⁻⁷ T. m/A

I₁ = current in conductor one=10 A

I₂ = current in conductor two= 10 A

l= length of conductor= 3 m

d= distance between the conductors = 5 mm = 0.005 m

Applying the values in the equation

F= 4π×10⁻⁷ *10*10*3 / 2π*0.005

F= 6 * 10⁻⁵ / 0.005

F=0.012 N

b

explanation:

bc its just b

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg