What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.30
The correct answer and explanation is:
To determine the magnitude of the rate of change of current that produces a self-induced emf of 6.30 V, use the formula for self-induced emf in an inductor: emf=LdIdt\text{emf} = L \frac{dI}{dt}
Where:
- emf is the self-induced electromotive force (in volts)
- L is the inductance of the coil (in henrys, H)
- dI/dt is the rate of change of current (in amperes per second, A/s)
If the problem provides an inductance value LL, substitute it and the emf value into the equation to solve for dIdt\frac{dI}{dt}.
Example:
If the inductance L=0.350 HL = 0.350 \, \text{H} and the self-induced emf is 6.30 V6.30 \, \text{V}, the formula becomes: 6.30=0.350⋅dIdt6.30 = 0.350 \cdot \frac{dI}{dt}
Solve for dIdt\frac{dI}{dt}: dIdt=6.300.350=18.0 A/s\frac{dI}{dt} = \frac{6.30}{0.350} = 18.0 \, \text{A/s}
Final Answer:
The magnitude of the rate of change of the current must be 18.0 A/s.
Explanation:
Self-induction is the process by which a changing current in a coil produces an emf in the same coil. This emf opposes the change in current due to Lenz’s Law. The faster the current changes, the greater the emf produced.
The inductance LL is a measure of how much opposition the coil offers to the change in current. A larger inductance results in a greater emf for the same rate of current change.
In this example, to produce a 6.30-volt self-induced emf in a 0.350-henry inductor, the current must be changing at a rate of 18.0 amperes per second. This relationship is linear, meaning that if the emf or inductance changes, the required rate of current change adjusts proportionally. This principle is foundational in electronics, particularly in circuits involving transformers, motors, and inductors.