Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
I think its all four of them could be wrong but try all four !!!!!!
Answer:
4.54
Explanation:
X+10X=50
11X=50
X=4.54#
<h2>please follow me...</h2>
Answer:
a = 1.764m/s^2
Explanation:
By Newton's second law, the net force is F = ma.
The equation for friction is F(k) = F(n) * μ.
In this case, the normal force is simply F(n) = mg due to no other external forces being specified
F(n) = mg = 15kg * 9.8 m/s^2 = 147N.
F(k) = F(n) * μ = 147N * 0.18 = 26.46N.
Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.
Thus, F(net) = F(k) = ma
26.46N = 15kg * a
a = 1.764m/s^2
Answer:
Given,
mass of man = 100 N = 10 kg
height = h = 25m
since the man does not move anything with his force, work done by him is zero
work done on the man = gain in potential energy
P.E=mgh
P.E=10×9.8×25
P.E=2.45KJ
Explanation:
so, potential energy gained by man is 2.45 KJ