In the given figure magnetic field at the centre of ring (O) is $10 \sqrt{2} T .$ Now it is bent through 90^o about XX' axis, so that two semicircular parts are mutually perpendicular. Then find the value of magnetic field (in Tesla) at centre:

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$10 \sqrt{2}$

$5 \sqrt{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} B = \frac{μ_{0} i}{2 R} B_{net} = \sqrt{B_{1}^{2} + B_{2}^{2}} \\ = 10 \sqrt{2} T and \\ B_{1} = B 2 = \frac{B}{2} = 5 \sqrt{2} T \\ ∴ B_{Net} = \sqrt{(5 \sqrt{2})^{2} + (5 \sqrt{2})^{2}} \\ = 10 T \end{array}$

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