The figure shows the cross section of a long cylindrical conductor through which an axial hole of radius $r$ is drilled with its centre at point $A$ . $O$ is the centre of the conductor. If an identical hole were to be drilled centred at point $B$ while maintaining the same current density the magnitude of magnetic field at $O$
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Initially $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{s}}$
Finally
$\mathrm{B}_{2}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{1}^{2}+2 \mathrm{B}_{1} \mathrm{B}_{2} \cos 120^{\circ}}$
$=2 \mathrm{B}_{1} \cos 60^{\circ}=\mathrm{B}_{1}$
So, $B$ remains unchanged.
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