Upon the insertion of an iron core with a cross-sectional area of 4 cm^2 , the magnetic field inside a long solenoid having 50 turns cm^ -1 increases from 2.5 × 10^ -3 T to 2.5 T. Determine the current flowing in the

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Correct answer: D. 0.4 A

The magnetic field inside a long solenoid without a core is given by

B_0 = \mu_0 n I

.

Given:
Initial magnetic field,

B_0 = 2.5 \times 10^{-3} \text{ T}

Number of turns per unit length,

n = 50 \text{ cm}^{-1} = 5000 \text{ m}^{-1}

Permeability of free space,

\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}

Substituting the values into the formula:

2.5 \times 10^{-3} = 4\pi \times 10^{-7} \times 5000 \times I

2.5 \times 10^{-3} = 20000\pi \times 10^{-7} \times I

2.5 \times 10^{-3} = 2\pi \times 10^{-3} \times I

I = \dfrac{2.5}{2\pi} \approx \dfrac{2.5}{6.28} \approx 0.4 \text{ A}

Upon the insertion of an iron core with a cross-sectional area of 4...

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