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:

When performing calculations with induced emf, it is safe to stick to absolute magnitudes to avoid confusion with signs: \[ L = \frac{|\varepsilon|}{\left|\frac{\Delta I}{\Delta t}\right|} \] Here, the current changes by 6 Amperes in 0.6 seconds, giving a rate of change of \(\frac{6}{0.6} = 10 \, \text{A/s}\). This simple step simplifies the calculation to: \(L = \frac{50}{10} = 5 \, \text{H}\).

Answer 1

Concept: Self-inductance (\(L\)) is a measure of a coil's resistance to changes in electric current. It acts as electrical inertia.
• Induced Emf Formula: The average magnitude of the electromotive force induced in an inductive system depends on the self-inductance and the average rate of change of current: \[ \varepsilon = -L \frac{\Delta I}{\Delta t} \quad \Rightarrow \quad |\varepsilon| = L \left| \frac{\Delta I}{\Delta t} \right| \] Where \(\Delta I = I_{\text{final}} - I_{\text{initial}}\) is the net difference in current amplitude, and \(\Delta t\) is the specific duration of the time interval.
• Dimensional Analysis: We can determine the dimensions of \(L\) by isolating it from the energy stored in an inductor: \[ U = \frac{1}{2} L I^2 \quad \Rightarrow \quad L = \frac{2U}{I^2} \] Since energy (\(U\)) has dimensions of work done, and current (\(I\)) has basic fundamental dimensions of Amperes (\(A\)), we can evaluate the dimensional structure.

Step 1: Deriving the Dimensional Formula of Self-Inductance (\(L\)).
Let us use the fundamental definition relating magnetic energy stored in an inductor: \[ \text{Energy } (U) = \frac{1}{2} L I^2 \] Isolating the self-inductance term: \[ L = \frac{2U}{I^2} \] Now, substituting the standard dimensional formulas for each constituent parameter:
• Dimensional formula of Potential/Kinetic Energy or Work Done: \[ [U] = [Work] = [Force \times Displacement] = [M^1 L^1 T^{-2} \times L^1] = [M^1 L^2 T^{-2}] \]
• Dimensional formula of Electric Current: \[ [I] = [A^1] \quad \Rightarrow \quad [I^2] = [A^2] \] Combining these base units together in the expression for \(L\): \[ [L] = \frac{[M^1 L^2 T^{-2}]}{[A^2]} = [M^1 L^2 T^{-2} A^{-2}] \] Answer Part 1: The dimensional formula for self-inductance is \([M^1 L^2 T^{-2} A^{-2}]\).

Step 2: Extracting numerical data from the text for the mathematical problem.
From the problem statement, we have the following values:
• Initial value of current, \(I_1 = 8.0 \, \text{A}\)
• Final value of current, \(I_2 = 2.0 \, \text{A}\)
• Continuous time interval, \(\Delta t = 0.6 \, \text{s}\)
• Average induced electromotive force amplitude, \(\varepsilon = 50 \, \text{V}\)

Step 3: Calculate the change in current (\(\Delta I\)) and the self-inductance \(L\).
The net change in current traversing the coil is: \[ \Delta I = I_2 - I_1 = 2.0 \, \text{A} - 8.0 \, \text{A} = -6.0 \, \text{A} \] Now, utilizing Faraday's self-induction formula to relate the given values: \[ \varepsilon = -L \frac{\Delta I}{\Delta t} \] Substitute the values into the equation: \[ 50 = -L \left( \frac{-6.0}{0.6} \right) \] Simplifying the terms within the parentheses: \[ \frac{-6.0}{0.6} = -10 \] Substitute this back into the relation: \[ 50 = -L \times (-10) \] \[ 50 = 10L \] Isolating the variable \(L\) by dividing both sides by 10: \[ L = \frac{50}{10} = 5 \, \text{H} \] The SI unit for self-inductance is the Henry (H). Answer Part 2: The self-inductance of the given coil is \(5 \, \text{H}\) (Henries).