The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
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Answer:
The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field.
Explanation:
<span>A spring is water coming from under the ground to the surface of the earth and a stream is water that is running along the ground through a trench like place on earth down a hill or steep a area.</span>
Answer:
9.38 m/s
Explanation:
Mass is conserved.
m₁ = m₂
ρ₁ Q₁ = ρ₂ Q₂
Assuming no change in density:
Q₁ = Q₂
v₁ A₁ = v₂ A₂
v₁ π r₁² = v₂ π r₂²
v₁ r₁² = v₂ r₂²
Plugging in values:
(1.50 m/s) (0.0250 m)² = v (0.0100 m)²
v = 9.38 m/s
