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Two concentric coils each of radius equal to $2 π cm$ are placed at right angles to each other. $3$ ampere and $4$ ampere are the currents flowing in each coil respectively. The magnetic induction in weber $/ m^{2}$ at the centre of the coils will be $(μ_{0} = 4 π \times 10^{- 7} Wb / A . m)$

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$5 \times 10^{5}$

$7 \times 10^{5}$

$12 \times 10^{5}$

$10^{5}$

answer is A.

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Detailed Solution

$B = \sqrt{{B_{1}}^{2} + {B_{2}}^{2}}$

$= \sqrt{{[\frac{μ_{0} \times 3}{2 (2 π)}]}^{2} + {[\frac{μ_{0} \times 4}{2 (2 π)}]}^{2}} = 5 \times 10^{- 5} T$

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