An inductive coil is a passive electrical component that stores energy in a magnetic field when electric current flows through it. The ability of a coil to store energy in a magnetic field is quantified by its inductance, measured in Henry (H).

Inductive Coil and Counter Voltage

When the current flowing through an inductive coil changes, it creates a changing magnetic flux. According to Faraday's Law of Electromagnetic Induction, this changing magnetic flux induces an electromotive force (EMF) or voltage across the coil. This induced voltage opposes the change in current that caused it, and it is often referred to as a counter voltage or back EMF.

The magnitude of this induced counter voltage (\(\varepsilon\)) is directly proportional to the inductance (\(L\)) of the coil and the rate at which the current (\(I\)) through it changes over time (\(\frac{dI}{dt}\)). The relationship is given by the formula:

\(\varepsilon = -L \frac{dI}{dt}\)

The negative sign indicates that the induced voltage opposes the change in current (Lenz's Law). However, when we are asked about the magnitude of the counter voltage or the rate change of current, we typically consider the absolute value:

\(\varepsilon = L \left| \frac{dI}{dt} \right|\)

Given Parameters for Current Rate Calculation

From the problem statement, we are provided with the following information about the inductive coil:

• Inductance of the coil (\(L\)) = 10 Henry (H)
• Counter voltage developed (\(\varepsilon\)) = 50 Volts (V)

We need to determine the rate change of current (\(\frac{dI}{dt}\)) in the coil.

Calculating the Rate Change of Current

To find the rate change of current, we can rearrange the formula \(\varepsilon = L \frac{dI}{dt}\) to solve for \(\frac{dI}{dt}\):

\(\frac{dI}{dt} = \frac{\varepsilon}{L}\)

Now, we can substitute the given values into this equation:

\(\frac{dI}{dt} = \frac{50 \, \text{Volts}}{10 \, \text{Henry}}\)

Performing the division:

\(\frac{dI}{dt} = 5 \, \text{Ampere/second}\)

The unit for the rate change of current is Ampere per second (A/s), which is consistent with our calculation.

Summary of Rate Change

Therefore, for an inductive coil with an inductance of 10 H that develops a counter voltage of 50 Volts, the rate change of current in the coil must be 5 Ampere/second.

This type of calculation is fundamental in understanding how inductors behave in electrical circuits, especially when dealing with transient phenomena where current levels are not constant.