Answer:
Naturally occurring radio waves are made by lightning or by astronomical objects. Artificially generated radio waves are used for fixed and mobile radio communication, broadcasting, radar and other navigation systems, communications satellites, computer networks and innumerable other applications.
Explanation:
Answer: Jupiter's mass
Explanation:
From Kepler's third law:

where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.
If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

Answer:
yes you are totally right
Answer:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH =
Current in the coil, i =
A
where

Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:

![e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20L%5Cfrac%7Bd%7D%7Bdt%7D%281680%29cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
![e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20-%207.50%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B0.0250%7D%5Ctimes%20%5Cfrac%7Bd%7D%7Bdt%7D%281680%29sin%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
For maximum emf,
should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:

(b) To calculate the maximum average flux,we know that:

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:


Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =