Complete question:
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.
Answer:
The bottom current is 12.8 A to the right.
Explanation:
Given;
length of the wires, L = 3.0 m
current in the top wire, I₁ = 12.5 A
repulsive force between the two wires, F = 2.4 x 10⁻⁴ N
distance between the two wires, r = 40 cm = 0.4 m
The repulsive force between the two wires is given by;

Where;
I₂ is the bottom current
The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

Therefore, the bottom current is 12.8 A to the right.
If your current exam mean is 97.2. and you receive a 99 on the next exam, then this will have the effect of increasing the mean.
<h3>What is the mean?</h3>
In statistics, the mean is an average value used to calculate when taking different measurements, which can be fundamental to collecting statistically significant information.
In conclusion, if your current exam mean is 97.2. and you receive a 99 on the next exam, then this will have the effect of increasing the mean.
Learn more about the average mean here:
brainly.com/question/20118982
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Answer:
The magnitude of the electric flux is 
Explanation:
Given that,
Electric field = 2.35 V/m
Angle = 25.0°
Area 
We need to calculate the flux
Using formula of the magnetic flux


Where,
A = area
E = electric field
Put the value into the formula



Hence, The magnitude of the electric flux is 