Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil
Answer:
6.169 μA
Explanation:
Formula for induced EMF is given by the equation;
EMF = M(di/dt). We are given;
di/dt = 2.5 A/s
M = 19μH = 19 × 10^(-6) H
Thus;
EMF = 19 × 10^(-6) × 2.5.
EMF = 47.5 × 10^(-6) V
Formula for current is;
i = EMF/R. R is resistance given as 7.7 ohms.
Thus; i = 47.5 × 10^(-6)/7.7
i = 6.169 μA
The line charge E-field Ec = λ/(2πr*e0),
where λ = charge/length and e0 is the permittivity constant = 8.8542E-12 F/m.
<span>The point charge E-field Ep = kq/r^2 where electrostatic constant k = 1/(4π*e0) = 8.99E9 N-m^2/C^2.</span>
The the object that has the less mass will travel faster because let’s say for example the first object has the mass of M and the second object has the mass of 2M and if the momentum is equal so that means that we could divide M with M and we would get V1=2V2 (being V1 the velocity of the first mass and V2 the velocity of the second one) I hope I helped you out.
An ampere (AM-pir), or amp
