The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current ' I ' is 1.57 × 10⁻⁶ Wb. Its magnetic moment will be [μ₀ = 4π × 10⁻⁷, SI unit and cross-sectional area is very small as compared to length of solenoid.] (a) 1 A m² (b) 0.25 A m² (c) 0.5 A m² (d) 0.75 A m²

Answer: (d). Φ = (μ₀ NIA)/(l) ⇒ M = NIA = (Φ l)/(μ₀). M = 1.57×10⁻⁶ × 0.64π×10⁻⁷ = 0.942×10⁻⁶1.256×10⁻⁶ = 0.75 So the answer is (d) 0.75 A m ².

The magnetic flux near the axis and inside the air core solenoid of… | MHT CET 2021 (21 Sep, Shift 2)

Q55MHT CET 2021 · 21 SEP · SHIFT 2Medium

Current Loop as a Magnetic Dipole

The magnetic flux near the axis and inside the air core solenoid of length $60\text{ cm}$ carrying current ' $I$ ' is $1.57 \times 10^{-6}\text{ Wb}$ . Its magnetic moment will be $[\mu_0 = 4\pi \times 10^{-7}$ , SI unit and cross-sectional area is very small as compared to length of solenoid.]

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Answer: (d)

$\Phi = \frac{\mu_0 NIA}{l} \Rightarrow M = NIA = \frac{\Phi l}{\mu_0}$ .
$M = \frac{1.57\times10^{-6} \times 0.6}{4\pi\times10^{-7}} = \frac{0.942\times10^{-6}}{1.256\times10^{-6}} = 0.75$
So the answer is (d) $0.75$ A m $^2$ .

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