A long straight rod of diameter 4 text{ mm} c | vedclass

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(D) The radius of the rod is $R = 	ext{diameter} / 2 = 4 	ext{ mm} / 2 = 2 	ext{ mm}$.For a point inside the rod at distance $r_1 = 1 	ext{ mm}$ $

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$1: 1$

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(D) The radius of the rod is $R = 	ext{diameter} / 2 = 4 	ext{ mm} / 2 = 2 	ext{ mm}$.For a point inside the rod at distance $r_1 = 1 	ext{ mm}$ $(r_1 $B_1 = \frac{\mu_0 i r_1}{2 \pi R^2}$For a point outside the rod at distance $r_2 = 4 	ext{ mm}$ $(r_2 > R)$,the magnetic field is given by:$B_2 = \frac{\mu_0 i}{2 \pi r_2}$Taking the ratio of the magnetic fields:$\frac{B_1}{B_2} = \frac{\frac{\mu_0 i r_1}{2 \pi R^2}}{\frac{\mu_0 i}{2 \pi r_2}} = \frac{r_1 r_2}{R^2}$Substituting the values $r_1 = 1 	ext{ mm}$,$r_2 = 4 	ext{ mm}$,and $R = 2 	ext{ mm}$:$\frac{B_1}{B_2} = \frac{1 	imes 4}{(2)^2} = \frac{4}{4} = 1: 1$

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

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