The figure shows the plot of magnitude of induced emf (\( \varepsilon \)) versus the rate of change of current in two coils ‘1’ and ‘2’. Which coil has a greater value of self-inductance and why?
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The self-inductance of a coil can be determined from the slope of the \( \varepsilon \) versus \( \frac{dI}{dt} \) plot. A steeper slope indicates a higher value of self-inductance.
The induced electromotive force (emf) in a coil is related to the rate of change of current through the coil by Faraday's law of induction, which is given by:
\[
\varepsilon = L \frac{dI}{dt}
\]
where:
- \( \varepsilon \) is the induced emf,
- \( L \) is the self-inductance of the coil,
- \( \frac{dI}{dt} \) is the rate of change of current.
From this equation, we see that the induced emf (\( \varepsilon \)) is directly proportional to the rate of change of current (\( \frac{dI}{dt} \)) and the self-inductance \( L \) of the coil. Therefore, the coil with the steeper slope in the plot will have a greater value of self-inductance.
Analyzing the Plot:
- In the plot, coil 1 has a steeper slope than coil 2, which means that for the same rate of change of current (\( \frac{dI}{dt} \)), coil 1 produces a larger induced emf.
- According to the relationship \( \varepsilon = L \frac{dI}{dt} \), a larger induced emf for the same rate of change of current implies a larger value of self-inductance.
Thus, coil 1 has a greater value of self-inductance than coil 2.
