The emf induced in the coil is -5.65 V
<h3>Induced emf in coil</h3>
The induced emf in the coil is given by ε = -NΔΦ/Δt where
- ΔΦ = change in magnetic flux Φ₂ - Φ₁ where
- Φ₁ = initial magnetic flux = -58 Wb and
- Φ₂ = final magnetic flux = 38 Wb and and
- Δt = change in time = t₂ - t₁ where
- t₁ = initial time = 0 s and
- t₂ = final time = 34 sand
- N = number of loops of coil = 2
Since ε = -NΔΦ/Δt
ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)
Substituting the values of the variables into the equation, we have
ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)
ε = -2(38 Wb - (-58 Wb))/(34 s - 0 s)
ε = -2(38 Wb + 58 Wb)/(34 s - 0 s)
ε = -2(96 Wb)/34 s
ε = -192 Wb/34 s
ε = -5.65 Wb/s
ε = -5.65 V
So, the emf induced in the coil is -5.65 V
Learn more about induced emf in coil here:
brainly.com/question/13051297
To develop this problem it is necessary to apply the concepts performed to the absolute pressure based on the reference pressure (atmospheric) and the pressure that is generated due to the height of the column of the measured liquid.
In mathematical terms the previous concept can be expressed as
Where
Atmospheric Pressure
Density
g = Gravitational acceleration
h = Height
Our values are given as
g = 9.8m/s
Replacing we have then that
Therefore the absolute pressure in the test section is 99.9019kPa
To calculate charge(Q) through a point, we multiply the current(I) with time (t)
Thus, Q= It
Therefore, t must be in SI units which gives 60*10 seconds.
Q=0.25A*600s= 150 coulombs.
