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Due to straight current carrying conductor, a null point occurred at p on east of the conductor. The net magnetic induction at a point Q which is half the distance of ‘p’ on north of the conductor is

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$\sqrt{5} B_{H}$

$\sqrt{2} B_{H}$

B_H

answer is D.

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Detailed Solution

$\begin{array}{l} B_{H} = \frac{μ_{0} I}{2 πr}; B_{N} = \sqrt{B_{H}^{2} + B^{2}} \\ = \sqrt{B_{H}^{2} + {[\frac{μ_{0} I}{2 π (\frac{r}{2})}]}^{2}} = \sqrt{B_{H}^{2} + 4 B_{H}^{2}} = B_{H} \sqrt{5} \end{array}$

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