Answer:
a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H
Explanation:
a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.
Also, Li = Ф where L = inductance of solenoid.
So, Li = NAB
= NA(μ₀ni)
= NA(μ₀Ni/l)
Li = μ₀AN²i/l
dividing both sides by i, we have
So, L = μ₀AN²/l
b. The self- inductance, L = μ₀AN²/l where
A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m
So, L = μ₀AN²/l
L = μ₀πd²N²/4l
L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)
L = 11,103.3 × 10⁻¹¹ H
L = 1.11033 × 10⁻⁷ H
L ≅ 1.11 × 10⁻⁷ H
Answer:
Exophthalmos
Explanation:
Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.
It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.
It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.
Answer:
3.0 cm
Explanation:
We can solve this problem by using the mirror equation:
where
f is the focal length of the mirror
p is the distance of the object from the mirror
q is the distance of the image from the mirror
In this problem we have:
f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)
p = 3.0 cm is the distance of the object from the mirror
Therefore, the distance of the image is:
And the positive sign means that the image is real.
(The second part of the exercise is just the description of the image of the first exercise).
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
Therefore, the number of turns of wire needed is 573.8 turns
