A current carrying circular coil of radius 11 cm is placed with its plane in magnetic meridian. The resultant magnetic induction at the centre of the coil is [Given current in the coil is 7A and B_H=0.4 Gauss ]

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$4 \sqrt{2} \times 10^{- 5} T$

$4 \times 10^{- 5} T$

$3 \times 10^{- 5} T$

answer is C.

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Detailed Solution

$B_{c} = \frac{μ_{0} i}{2 r} = \frac{4 \times \frac{22}{7} \times 10^{- 7} \times 7}{2 \times 11 \times 10^{- 2}} = 0 .4 G$
$∴ B_{Net} = \sqrt{B^{2} + B^{2}} = \sqrt{2} \cdot B = \sqrt{2} \times 0 .4 G = 4 \sqrt{2} \times 10^{5} T$

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