Its either releasing a stretch rubber band and a ball rolling down a hill because rubber have a potential to move when its stretch and its an eleastic potential energy also a ball rolling down a hill is a gravitational to kinetic energy because the ball started at rest which is art the top of the hill and it has a potential to fall. Once its started rolling its energy transfer into kinetic energy.
Hope this helps
Answer:
average emf induced in the coil is 1.57 x V
Explanation:
Given data
n = 10 turns per centimeter = 1000 m^-1
N = 5
cross-sectional area A = 5.0 cm2 = 5.0 x 10^-4 m²
current = 0.25 A
to find out
average emf induced in the coil
solution
we will find emf by the given formula
emf = - (μ×N×n×A×ΔI ) / t ..................1
here
μ = 4π x 10^-7 and
N = 10
n = 1000
A = 5.0 x 10^-4
ΔI = 0.25
t = 0.050
put all value in equation 1
emf = - (μ×N×n×A×ΔI ) / t
emf = - (4π x 10^-7 ×5 ×1000× 5.0 x 10^-4 × 0.25 ) / 0.050
emf = 1.57 x V
average emf induced in the coil is 1.57 x V
Explanation:
Given that,
Induced emf
Rate of current = 0.250 A/s
Number of turns =30
(a). We need to calculate the mutual inductance of the pair of coils
Using formula of the mutual inductance
The mutual inductance of the pair of coils is 6.6 mHz.
(b). We need to calculate the flux through each turn
Using formula of flux
Put the value into the formula
The flux through each turn is
(c). We need to calculate the magnitude of the induced emf in the first coil
Using formula of induced emf
The magnitude of the induced emf in the first coil is
Hence, This is the required solution.