Solution: Current in a Coil upon Core Removal
Step 1: Analyze the Graph
The graph plots Inductance ($L$) versus Displacement ($x$). We need to read the values for the initial and final states.
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Initial State ($x=0$): The core is fully inserted. The graph shows a plateau.
Looking at the Y-axis, the plateau is exactly halfway between 4.0 and 6.0 mH.
$$ L_1 = 5.0 \text{ mH} $$ -
Final State (Core removed): When $x$ is large (core is out), the inductance drops to a base level.
The graph shows this level is halfway between 0.0 and 2.0 mH.
$$ L_2 = 1.0 \text{ mH} $$
Step 2: Physics Principle
The problem states the core is taken out “quickly”. This implies the time interval $\Delta t \to 0$. Due to the property of inductance (Faraday’s Law), the magnetic flux through the coil cannot change instantaneously.
Therefore, Magnetic Flux ($\Phi$) is conserved:
$$ \Phi_{\text{initial}} = \Phi_{\text{final}} $$
$$ L_1 I_1 = L_2 I_2 $$
Step 3: Calculation
We substitute the known values:
$$ (5.0 \text{ mH}) \times (1.0 \text{ A}) = (1.0 \text{ mH}) \times I_2 $$
Solving for the final current $I_2$:
$$ I_2 = \frac{5.0 \times 1.0}{1.0} $$
$$ I_2 = 5.0 \text{ A} $$
Final Answer: The current in the coil is 5.0 A.