Question

Write the dimensional formula for self-inductance. The current in a coil changes from 8.0 A to 2.0 A in 0.6 s. If an average emf induced in the coil is 50 V, calculate the self-inductance of the coil.

Hint:

Using the standard magnetic energy formula $U = \frac{1}{2}LI^2$ is frequently the absolute fastest, most reliably error-free algebraic pathway to reliably deriving the often-forgotten dimensional formula for inductance.
When meticulously calculating absolute physical properties strictly like inductance or resistance, strictly using the absolute scalar magnitude of the dynamic current change safely and effectively prevents horribly confusing negative results.

Answer 1

Concept:
• Self-inductance firmly and mathematically measures a conductive coil's inherent natural ability to dynamically oppose any rapid, sudden changes in its internal flowing electrical current.

• Dimensional formulas elegantly and systematically break down complex, derived electromagnetic quantities strictly into their fundamental baseline mechanical and electrical base units.

• The induced electromotive force strictly links a coil's self-inductance mathematically with the precise time-rate of temporal current change occurring within it.

Step 1: Derive the Dimensional Formula for Self-Inductance
We start strongly and strategically from the well-known, foundational energy formula involving an energized inductor:
\[ U = \frac{1}{2} L I^2 \]
Here, $U$ physically represents the stored magnetic potential energy, $L$ critically represents the self-inductance, and $I$ rigidly represents the operating current.
We strategically and algebraically rearrange this specific formula to clearly isolate the self-inductance variable on one side:
\[ L = \frac{2U}{I^2} \]
Since the numerical mathematical constant $2$ is completely dimensionless, the final physical dimension of $L$ entirely depends strictly on the fundamental physical dimensions of energy and current.
The universally accepted, standard dimensional formula for standard energy (equivalent to Work) is $[ML^2T^{-2}]$.
The firmly established standard dimensional formula for foundational electrical current is inherently $[A]$.
Substituting these specific reliable base dimensions smoothly and carefully into the algebraically rearranged equation yields:
\[ [L] = \frac{[ML^2T^{-2}]}{[A^2]} \]
Moving the current dimension cleanly into the upper numerator gives the final expression:
\[ [L] = [ML^2T^{-2}A^{-2}] \]

Step 2: Identify Parameters for Numerical Calculation
The initial electrical current strongly flowing in the coil is strictly recorded as $I_1 = 8.0 \text{ A}$.
The drastically and rapidly reduced final current is firmly recorded as $I_2 = 2.0 \text{ A}$.
The exact time interval over which this rapid, dynamic change thoroughly occurs is given as $\Delta t = 0.6 \text{ s}$.
The absolute average magnitude of the opposing induced electromotive force securely registered during this event is $|e| = 50 \text{ V}$.

Step 3: Calculate the Exact Self-Inductance
The primary governing mathematical formula firmly connecting all these dynamic variables is famously given by Faraday's and Lenz's laws combined together:
\[ |e| = L \left| \frac{\Delta I}{\Delta t} \right| \]
We mathematically and carefully determine the absolute magnitude of the rapid change in the operating current:
\[ |\Delta I| = |I_2 - I_1| = |2.0 - 8.0| = |-6.0| = 6.0 \text{ A} \]
Now, we meticulously substitute all the carefully collected and verified numerical values squarely back into the primary algebraic formula:
\[ 50 = L \times \left( \frac{6.0}{0.6} \right) \]
Calculating the simple internal mathematical fraction cleanly provides a very neat factor of exactly ten:
\[ 50 = L \times 10 \]
Isolating the self-inductance variable firmly by performing basic division elegantly yields the final answer:
\[ L = \frac{50}{10} = 5 \text{ H} \]
The total self-inductance of the rigorously tested induction coil is precisely evaluated to be strictly 5 Henrys.