Question

What is meant by self-inductance ? The self-inductance of a coil is 0.4 m Henry. The value of current flowing in it changes by 1 ampere in 0.1 second. Calculate the induced electro motive force.

Hint:

Pay close attention to units. Self-inductance is often given in millihenries (mH) or microhenries (\(\mu\)H). Always convert these to the base unit of Henry (H) before performing calculations to ensure the final answer is in Volts.

Answer 1

Step 1: Definition of Self-Inductance:
Self-inductance is the property of an electrical conductor or coil by which a change in the current flowing through it induces an electromotive force (e.m.f.) in the conductor itself. This induced e.m.f., often called a "back e.m.f.", always opposes the change in current that produces it (Lenz's Law). The magnetic flux (\(\Phi_B\)) linked with a coil is directly proportional to the current (I) flowing through it, so \( \Phi_B \propto I \). This can be written as \( \Phi_B = LI \), where L is the constant of proportionality called the self-inductance of the coil. The SI unit of self-inductance is the Henry (H).
Step 2: Key Formula for Calculation:
According to Faraday's law of electromagnetic induction, the induced e.m.f. (\(\epsilon\)) in a coil is equal to the negative rate of change of magnetic flux linked with it. \[ \epsilon = - \frac{d\Phi_B}{dt} \] Substituting \( \Phi_B = LI \), we get: \[ \epsilon = - \frac{d(LI)}{dt} \] Since L is a constant for a given coil, the formula becomes: \[ \epsilon = -L \frac{dI}{dt} \] where \( \frac{dI}{dt} \) is the rate of change of current.
Step 3: Detailed Calculation:
We are given the following values:
Self-inductance, \( L = 0.4 \, \text{mH} = 0.4 \times 10^{-3} \, \text{H} \)
Change in current, \( \Delta I = 1 \, \text{A} \)
Time interval, \( \Delta t = 0.1 \, \text{s} \)
First, we find the rate of change of current: \[ \frac{dI}{dt} \approx \frac{\Delta I}{\Delta t} = \frac{1 \, \text{A}}{0.1 \, \text{s}} = 10 \, \text{A/s} \] Now, we can calculate the magnitude of the induced e.m.f. using the formula. The negative sign indicates the direction of the e.m.f. (opposition to the change), but the question asks for the value. \[ |\epsilon| = L \left| \frac{dI}{dt} \right| \] \[ |\epsilon| = (0.4 \times 10^{-3} \, \text{H}) \times (10 \, \text{A/s}) \] \[ |\epsilon| = 4.0 \times 10^{-3} \, \text{V} \] This is equal to 4 millivolts (mV).
Step 4: Final Answer:
The induced electromotive force in the coil is \( 4 \times 10^{-3} \) volts or 4 mV.