A solenoidal coil with 25 turns of wire is wound tightly around another coil with 310 turns. the inner solenoid is 25.0 cm long
and has a diameter of 2.00 cm. at a certain time, the current in the inner solenoid is 0.116 a and is increasing at a rate of 1.85 103 a/s. (a) for this time, calculate the average magnetic flux through each turn of the inner solenoid. wb (b) for this time, calculate the mutual inductance of the two solenoids. h (c) for this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid. v
1 answer:
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Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
