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(D) For a long straight wire of radius $a$ carrying a uniform current $I$:$1$. Inside the wire $(r < a)$, the magnetic field $B_{in}$ is given by $B_{

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$3: 4$

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(D) For a long straight wire of radius $a$ carrying a uniform current $I$:$1$. Inside the wire $(r $2$. Outside the wire $(r > a)$, the magnetic field $B_{out}$ is given by $B_{out} = \frac{\mu_0 I}{2 \pi r}$.$3$. At $r_1 = 0.5a$ (inside), $B_1 = \frac{\mu_0 I (0.5a)}{2 \pi a^2} = \frac{\mu_0 I}{4 \pi a}$.$4$. At $r_2 = 1.5a$ (outside), $B_2 = \frac{\mu_0 I}{2 \pi (1.5a)} = \frac{\mu_0 I}{3 \pi a}$.$5$. The ratio is $\frac{B_1}{B_2} = \frac{\mu_0 I / 4 \pi a}{\mu_0 I / 3 \pi a} = \frac{3}{4}$.

$A$ long straight wire of circular cross-section of radius $a$ is carrying a steady current. The current is distributed uniformly across the cross-section of the wire. The ratio of the magnetic fields at points $0.5a$ and $1.5a$ from the centre of the wire is

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