The magnetic induction at a point P which is $4 c m$ distant from a long current-carrying wire is $10^{- 3} T .$ The field of induction at a distance $12 c m$ from the same current would be?

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Solution

Step 1: Given data.
The distance $r_{1} = 4 c m$
The magnetic induction at $r_{1}$ due to the current carrying wire is $B_{1} = 10^{- 3} T .$
The modified distance $r_{2} = 12 c m$
The magnetic induction at $r_{2}$ is $B_{2}$ (assume)
Step 2: Solve for magnetic induction $B_{2}$ :
The magnetic induction due to a current carrying straight wire at a point is inversely proportional to the distance of the point from the wire.
Therefore, $B α \frac{1}{r}$
$\frac{B_{1}}{B_{2}} = \frac{r_{2}}{r_{1}} B_{2} = B_{1} \times \frac{r_{1}}{r_{2}} B_{2} = 10^{- 3} \times \frac{4}{12} B_{2} = 3.33 \times 10^{- 4} T$
Hence, the required value of magnetic induction is $3.33 \times 10^{- 4}$

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