A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, the magnitude of the magnetic field inside the solenoid (near the centre) is:

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This question was previously asked in

AAI ATC Junior Executive 21 Feb 2023 Shift 2 Official Paper

Answer 1

Given:

Length of the solenoid, l = 80 cm = 0.8 m

Number of turns N = 400

Diameter of the solenoid, D = 1.8 cm = 0.018 m

Current carried by the solenoid, I = 8.0 A

Permeability \(\mu_o=4\pi \times10^{-7}TmA^{-1}\)

Concept:

Magnitude of the magnetic field inside the solenoid near its center is given by the relation:

\(B=\frac{\mu_0 NI}{l}\)

Explanation:

Number of Total turns in the solenoid N = \(5\times400=2000\ turns\)

\(B=\frac{\mu_0 NI}{l}\) ⇒ \(B=\frac{4\pi\times 10^{-7}\times2000\times 8}{0.8}\)

\(B=2.5\times10^{-2}T\)

Hence, the correct answer is Option (4).