Two thin, infinitely long, parallel wires are lying on the ground a distance d=3cm apart. They carry a current Io=200A going into the page. A third thin, infinitely long wire with mass per unit length λ=5g/m carries current I going out of the page. What is the value of the current I in amps in this third wire if it is levitated above the first two wires at height h=10cm above them and at a horizontal position midway between them?
draw the forces, there's repulsion between the two equal wire against the third one
We have:
F = [uo.I1.I2/(2.pi.d)]*Ltotal
2*F*sin(a)= m*g = lamba*Ltotal*g
sin(a)=h/0.5d
Replace the values and you'll I.
By the way, have you done the one with the thick metal slab?
have you got the answer?
hai anonymous ,i stuck on dis question if u done for it plz help ???
To find the value of the current I in the third wire, we can use Ampere's law. Ampere's law states that the integral of the magnetic field along a closed path is equal to the product of the current enclosed by the path and the permeability of free space (μ₀).
Here's how we can approach the problem step by step:
Step 1: Calculate the magnetic field produced by each of the parallel wires at the location of the third wire.
- The magnetic field produced by an infinitely long straight wire at a distance r from the wire is given by:
B = (μ₀ * I) / (2π * r)
Where B is the magnetic field, I is the current, μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A), and r is the distance from the wire.
- So, using this formula, we can calculate the magnetic field produced by each of the parallel wires at the location of the third wire.
Let's call the distance between the third wire and one of the parallel wires as r₁, and the distance between the third wire and the other parallel wire as r₂.
B₁ = (μ₀ * Io) / (2π * r₁)
B₂ = (μ₀ * Io) / (2π * r₂)
Step 2: Calculate the magnetic field due to the two parallel wires at the location of the third wire.
- Since the two parallel wires are carrying current in the same direction, the magnetic fields due to each wire will add up.
B_total = B₁ + B₂
Step 3: Calculate the magnetic force on the third wire.
- The magnetic force acting on a current-carrying wire in a magnetic field is given by:
F = I * L * B
Where F is the force, I is the current, L is the length of the wire, and B is the magnetic field.
- In this case, the force on the third wire is acting in the opposite direction of gravity and is responsible for levitating the wire. So, we can equate the magnetic force to the force due to gravity:
I * L * B_total = λ * L * g
Where λ is the mass per unit length of the wire and g is the acceleration due to gravity.
Step 4: Solve for the current I in the third wire.
- Rearranging the equation from Step 3, we can solve for the current I.
I = (λ * g) / B_total
Substituting the values of λ, g, and B_total into the equation will give us the value of I.
So, by following these steps, we can find the value of the current I in the third wire.