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.
Answer:

Explanation:
We can use Newton's Universal Law of Gravitation to solve this problem:
., where
is acceleration due to gravity at the planet's surface,
is gravitational constant
,
is the mass of the planet, and
is the radius of the planet.
Since acceleration due to gravity is given as
, our radius should be meters. Therefore, convert
kilometers to meters:
.
Now plugging in our values, we get:
,
Solving for
:
.
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