A long straight metal rod has a very long hole of radius ' $a$ ' drilled parallel to the rod axis as shown in the figure. If the rod carries a current $i,$ find the value of magnetic induction on the axis of the hole, where $O C = c$

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$\frac{μ_{0} i c}{π (b^{2} - a^{2})}$

$\frac{μ_{0} i c}{2 π (b^{2} - a^{2})}$

$\frac{μ_{0} i (b^{2} - a^{2})}{2 π c}$

$\frac{μ_{0} i c}{2 π a b}$

answer is B.

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Detailed Solution

Current per unit area $σ = \frac{i}{π (b^{2} - a^{2})}$

At C Magnetic field due to remaining portion =magnetic field due to whole portion – magnetic field due to cavity

$\Rightarrow$ $B = [\frac{μ_{0}}{2 π} \frac{I}{b^{2}}] c - 0$
Here $I = σ (π b^{2}) = \frac{i b^{2}}{(b^{2} - a^{2})}$

$∴ B = \frac{μ_{0} i c}{2 π (b^{2} - a^{2})}$

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