Step 1: Understanding the Concept:
Self-inductance (\(L\)) is a property of a coil by which it opposes a change in current flowing through it.
When a steady current is established in a circuit, work has to be done against the induced EMF.
This work is stored as magnetic potential energy in the inductor.
Step 2: Key Formula or Approach:
The magnetic energy stored (which is the work done, \(W\)) in an inductor carrying a current \(I\) is given by the formula:
\( W = \frac{1}{2}LI^2 \).
Step 3: Detailed Explanation:
From the energy formula, we can express self-inductance \( L \) in terms of work done \( W \) and current \( I \):
\[ L = \frac{2W}{I^2} \]
The question asks for a situation where self-inductance is numerically related to the work done.
Let's analyze the options based on a "unit current".
Assume a unit current is established in the circuit, so \( I = 1 \) unit.
Substitute \( I = 1 \) into the energy formula:
\[ W = \frac{1}{2}L(1)^2 = \frac{1}{2}L \]
Rearranging this to solve for \( L \):
\[ L = 2W \]
This means that numerically, the self-inductance \( L \) is equal to two times the work done \( W \) to establish a unit current.
Comparing this with the given options:
Option (A) is incomplete as it doesn't specify unit current.
Option (B) exactly matches our derivation: twice the work done for unit current.
Option (C) says thrice, which is incorrect.
Option (D) says equal to work done, which would mean \( L = W \), missing the factor of 2.
Step 4: Final Answer:
The self-inductance is numerically equal to twice the work done in establishing the magnetic flux associated with unit current.