Q2 A source of frequency 500 Hz emits waves of
1 answer:
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Answer:
(a) The current needed is 56.92 A
(b) The magnitude of the magnetic field inside the solenoid is 0.134 T
(c) The energy density inside the solenoid is 7.144 kJ/m³
Explanation:
Given;
energy stored in the magnetic field of solenoid, E = 8.1 J
number of turns of the solenoid, N = 620 turns
diameter of the solenoid, D = 6.6 cm = 0.066 m
radius of the solenoid, r = D/2 = 0.033 m
length of the solenoid, L = 33 cm = 0.33 m
Inductance of the solenoid is given as;
where;
A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
(A). How much current needed
Energy stored in magnetic field of solenoid is given as;
Where;
I is the current in the solenoid
(B) The magnitude of the magnetic field inside the solenoid
B = μ₀nI
where;
n is number of turns per unit length
B = μ₀(N/L)I
B = (4π x 10⁻⁷)(620/0.33)(56.92)
B = 0.134 T
(C) The energy density (energy/volume) inside the solenoid
The energy of a single photon is given by
where
is the Planck constant
f is the frequency of the wave (of the photon)
In our problem, the radio wave has a frequency of
, so if we put this value into the previous formula, we can find the energy of a single photon of this electromagnetic wave:
where
f is the frequency of the wave (of the photon)
In our problem, the radio wave has a frequency of