A coil has an inductance of 9.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find the magnitude of the average induced emf in the coil during this time interval.

Correction: E = 0.039 Volts.

To find the magnitude of the average induced emf in the coil, we can use the formula:

emf = L * ΔI/Δt

Where:
emf is the magnitude of the average induced emf
L is the inductance of the coil (9.00 mH)
ΔI is the change in current (ΔI = 1.50 A - 0.200 A = 1.30 A)
Δt is the change in time (Δt = 0.300 s)

Now let's substitute the given values into the formula:

emf = 9.00 mH * 1.30 A / 0.300 s

First, we need to convert millihenries to henries:

9.00 mH = 9.00 * 10^(-3) H

emf = (9.00 * 10^(-3) H) * 1.30 A / 0.300 s

Next, we can simplify the expression:

emf = (0.0117 H * A) / 0.300 s

emf = 0.0389 V

Therefore, the magnitude of the average induced emf in the coil during this time interval is 0.0389 V.

To find the magnitude of the average induced emf in the coil, we can use Faraday's law of electromagnetic induction, which states that the induced emf (ε) is equal to the rate of change of magnetic flux through the coil.

The formula for the average induced emf in a coil is given by:

ε = -L * (ΔI/Δt)

where ε is the induced emf, L is the inductance of the coil, ΔI is the change in current, and Δt is the time interval.

Given that the inductance of the coil (L) is 9.00 mH (millihenries), the change in current (ΔI) is (1.50 A - 0.200 A), and the time interval (Δt) is 0.300 s, we can substitute these values into the formula to calculate the average induced emf:

ε = -L * (ΔI/Δt)
= -(9.00 * 10^-3 H) * ((1.50 A - 0.200 A) / 0.300 s)

First, let's calculate the change in current:

ΔI = 1.50 A - 0.200 A
= 1.30 A

Now, substitute the values into the formula to calculate the average induced emf:

ε = -(9.00 * 10^-3 H) * (1.30 A / 0.300 s)

Next, simplify the expression:

ε = -(9.00 * 10^-3 H) * (4.33 A/s)
= -39.37 * 10^-3 V
= -39.37 mV

The magnitude of the average induced emf in the coil during this time interval is 39.37 millivolts (mV).

E = L*di/dt = (9*10^-3)*(1.5-0.2)/0.3 = 4.33 Volts.