The self-inductance of a coil will change by 8 times its original value by increasing its radius value by 2 and increasing the length of the coil by 2.
Self-Inductance: -
The definition of self-inductance is the induction of a voltage in a wire that carries current when the current in the wire is changing. In the instance of self-inductance, the circuit itself induces a voltage through the magnetic field produced by a changing current.
We know that the self-inductance of the coil is denoted by: -
L= µ *π*(r)^2*(N)^2*l
Where
L= Self-Inductance of the coil
µ= Magnetic Permeability Constant
r= Radius of the coil
l= Length of the coil
N= Number of turns of the coil
Here Self-inductance of the coil is directly proportional to the length of the coil and the square of the radius of the coil.
So,
On increasing the radius of the coil by a factor of 2 and the length of the coil by 2 the self-inductance of the coil increases by 8 times its original value.
Learn more about Self-Inductance here: -
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Elliptical and Spiral have some similarities, they both are huge and contain lost of dust and also they are held by gravitational forces.
340 ms
I got it right and I hope you do as well
Answer:
a) x = ⅔ d
, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
