The magnetic flux near the axis and inside the air core solenoid of length 80 cm carrying current ' I ' is 1.57 × 10⁻⁶ Wb. Its magnetic moment will be [cross-sectional area of a solenoid is very small as compared to its length, μ₀ = 4π × 10⁻⁷ SI unit ](π = 3.14) (a) 0.25 Am² (b) 0.50 Am² (c) 1 Am² (d) 1.2 Am²

Answer: (c). Flux inside: Φ = (μ₀ NIA)/(l), so M = NIA = (Φ l)/(μ₀). M = 1.57×10⁻⁶ × 0.84×3.14×10⁻⁷ = 1.256×10⁻⁶1.256×10⁻⁶ So the answer is (c) 1 A m ².

The magnetic flux near the axis and inside the air core solenoid of… | MHT CET 2024 (11 May, Shift 1)

Q32MHT CET 2024 · 11 MAY · SHIFT 1Medium

Current Loop as a Magnetic Dipole

The magnetic flux near the axis and inside the air core solenoid of length 80 cm carrying current ' I ' is $1.57 \times 10^{-6}\text{ Wb}$ . Its magnetic moment will be [cross-sectional area of a solenoid is very small as compared to its length, $\mu_0 = 4\pi \times 10^{-7}$ SI unit $](\pi = 3.14)$

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Answer: (c)

Flux inside: $\Phi = \frac{\mu_0 NIA}{l}$ , so $M = NIA = \frac{\Phi l}{\mu_0}$ .
$M = \frac{1.57\times10^{-6} \times 0.8}{4\times3.14\times10^{-7}} = \frac{1.256\times10^{-6}}{1.256\times10^{-6}}$
So the answer is (c) $1$ A m $^2$ .

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