Rt= ΣR = 40Ω
Vt= 80V
It= 80V/40Ω= 2A
V1= 15Ω*2A= 30V
V2= 20Ω*2A= 40V
V3= 5Ω*2A= 10V
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:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
So, the force has a magnitude of 4212 N
Hey
Potential Difference given is : 2V
Resistance is : 2 ohms
By Ohm's Law, one can easily utilize the relation :
Where, { v , i , r } are the potential difference, current and Resistance Respectively.
Hence,
Hence, the Current is 1 Ampere