The self inductance of solenoid of length 31.4 cm, area of cross section 10⁻³ m² having total number of turns 500 will be nearly [ μ₀ = 4π × 10⁻⁷ SI unit] (a) 3 × 10⁻⁶ H (b) 2 × 10⁻⁶ H (c) 0.5 × 10⁻⁶ H (d) 4 × 10⁻⁶ H

Answer: (b). L = (μ₀ N² A)/(l) with l = 0.314 m = 0.1π m. L = 4π×10⁻⁷×(500)²×10⁻³0.1π L = 4×10⁻⁷×2.5×10⁵×10⁻³0.1 = 1×10⁻³ H None of the options matches; the key gives (b) 2×10⁻⁶ H.

The self inductance of solenoid of length 31.4 cm, area of cross… | MHT CET 2021 (23 Sep, Shift 2)

Q67MHT CET 2021 · 23 SEP · SHIFT 2Easy

Self-Inductance & Energy

The self inductance of solenoid of length $31.4\text{ cm}$ , area of cross section $10^{-3}\text{ m}^{2}$ having total number of turns 500 will be nearly [ $\mu_0 = 4\pi \times 10^{-7}$ SI unit]

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Answer: (b)

$L = \dfrac{\mu_0 N^2 A}{l}$ with $l = 0.314\text{ m} = 0.1\pi\text{ m}$ .
$L = \dfrac{4\pi\times10^{-7}\times(500)^2\times10^{-3}}{0.1\pi}$
$L = \dfrac{4\times10^{-7}\times2.5\times10^{5}\times10^{-3}}{0.1} = 1\times10^{-3}\text{ H}$
None of the options matches; the key gives (b) $2\times10^{-6}\text{ H}$ .

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Self-Inductance & Energy question from MHT CET 2021 (23 Sep, Shift 2)

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