Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
<span>47 amu is the correct answer</span>
Milky Way Galaxy, same one as you.
Answer:
Explanation:
Given that,
Frequency of radio signal is
f = 800kHz = 800,000 Hz.
Distance from transmitter
d = 8.5km = 8500m
Electric field amplitude
E = 0.9 V/m
The average energy density can be calculated using
U_E = ½•ϵo•E²
Where ϵo = 8.85 × 10^-12 F/m
Then,
U_E = ½ × 8.85 × 10^-12 × 0.9²
U_E = 3.58 × 10^-12 J/m²
The average electromagnetic energy density is 3.58 × 10^-12 J/m²
