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Answer:
The magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.
Explanation:
Given;
distance between the parallel wires, r = 5.0 cm = 0.05 m
current in the first wire, I₁ = 1.65 A
current in the second wire, I₂ = 3.25 A
The magnitude of the force per unit length between the two wires is calculated as follows;
Therefore, the magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.
First of all, we need to write the First Law of thermodynamics assigning the correct sign convention:
where
is the change in internal energy of the system
Q is the heat absorbed/released
W is the work done
and the signs are assigned based on whether there is an increase in the internal energy or not. Therefore:
Q is positive if it is absorbed by the system (because internal energy increases)
Q is negative if it is released by the system (because internal energy decreases)
W is negative if it is done by the system on the surrounding (because internal energy decreases)
W is positive if it is done by the surrounding on the system (because internal energy increases)
Using these definitions, we can now fill the text of the question:
When a system is heated, heat is ABSORBED by the system. The amount of heat added is given a POSITIVE sign. When a system is cooled, heat is RELEASED by the system. The amount of heat is given a NEGATIVE sign. If a gas expands, it must push the surrounding atmosphere away. Thus, work is done BY the system and is given a NEGATIVE sign. If a gas is compressed, then work is done ON the system. This work is given a POSITIVE sign.