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(D) Given: Diameter $D = 4 \ mm$,so radius $R = 2 \ mm$. For a point inside the rod $(r < R)$,the magnetic field is given by $B_{in} = \frac{\mu_0 i r

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$1:1$

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(D) Given: Diameter $D = 4 \ mm$,so radius $R = 2 \ mm$. For a point inside the rod $(r At $r_1 = 1 \ mm$,$B_1 = \frac{\mu_0 i (1)}{2 \pi (2)^2} = \frac{\mu_0 i}{8 \pi}$. For a point outside the rod $(r > R)$,the magnetic field is given by $B_{out} = \frac{\mu_0 i}{2 \pi r}$. At $r_2 = 4 \ mm$,$B_2 = \frac{\mu_0 i}{2 \pi (4)} = \frac{\mu_0 i}{8 \pi}$. The ratio of the magnetic fields is $\frac{B_1}{B_2} = \frac{\mu_0 i / 8 \pi}{\mu_0 i / 8 \pi} = 1:1$.

$A$ long straight rod of diameter $4 \ mm$ carries a steady current '$i$'. The current is uniformly distributed across its cross-section. The ratio of the magnetic fields at distances $1 \ mm$ and $4 \ mm$ from the axis of the rod is

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