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(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 i}{2R}$.Given,$R = 2\pi \, cm = 2\pi 	imes 10^{-2} \, m$.For the

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$5 	imes 10^{-5}$

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(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 i}{2R}$.Given,$R = 2\pi \, cm = 2\pi 	imes 10^{-2} \, m$.For the first coil with current $i_1 = 3 \, A$:$B_1 = \frac{\mu_0 	imes 3}{2 	imes 2\pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 3}{4\pi 	imes 10^{-2}} = 3 	imes 10^{-5} \, T$.For the second coil with current $i_2 = 4 \, A$:$B_2 = \frac{\mu_0 	imes 4}{2 	imes 2\pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 4}{4\pi 	imes 10^{-2}} = 4 	imes 10^{-5} \, T$.Since the coils are placed at right angles,the resultant magnetic field is $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{(3 	imes 10^{-5})^2 + (4 	imes 10^{-5})^2} = \sqrt{9 + 16} 	imes 10^{-5} = \sqrt{25} 	imes 10^{-5} = 5 	imes 10^{-5} \, Wb/m^2$.

Two concentric coils each of radius equal to $2\pi \, cm$ are placed at right angles to each other. $3\, A$ and $4\, A$ are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^2$ at the centre of the coils will be $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$.

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