A long straight wire of radius ' r ' carries a steady current ' I '. The current is uniformly distributed over its cross-section. The ratio (B/B₁) of the magnetic field ' B ' and ' B₁ ' at radial distances ' r/2 ' and ' 3r ' respectively, from the axis of the wire is (a) 1/2 (b) 3/2 (c) 5/2 (d) 1/2

Answer: (b). At r/2 (inside): B = (μ₀ I (r/2))/(2π r²) = (μ₀ I)/(4π r). At 3r (outside): B₁ = (μ₀ I)/(2π (3r)) = (μ₀ I)/(6π r). B/B₁ = 6/4 = 3/2. So the answer is (b) 3/2.

A long straight wire of radius ' r ' carries a steady current ' I '… | MHT CET 2025 (25 Apr, Shift 1)

Q16MHT CET 2025 · 25 APR · SHIFT 1Medium

Ampere's Law, Cylinders, Solenoid & Toroid

A long straight wire of radius ' $r$ ' carries a steady current ' $I$ '. The current is uniformly distributed over its cross-section. The ratio $\left(\frac{B}{B_1}\right)$ of the magnetic field ' B ' and ' $\mathrm{B}_1$ ' at radial distances ' $\frac{r}{2}$ ' and ' $3r$ ' respectively, from the axis of the wire is

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Answer: (b)

At $r/2$ (inside): $B = \dfrac{\mu_0 I (r/2)}{2\pi r^2} = \dfrac{\mu_0 I}{4\pi r}$ .
At $3r$ (outside): $B_1 = \dfrac{\mu_0 I}{2\pi (3r)} = \dfrac{\mu_0 I}{6\pi r}$ .
$\dfrac{B}{B_1} = \dfrac{6}{4} = \dfrac{3}{2}$ .
So the answer is (b) $\dfrac{3}{2}$ .

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