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(A) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.Since the two coils are placed at right angles,their magn

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$5 	imes 10^{-5}$

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(A) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.Since the two coils are placed at right angles,their magnetic fields $B_1$ and $B_2$ will also be perpendicular to each other.The resultant magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2}$.Substituting $B_1 = \frac{\mu_0 i_1}{2r}$ and $B_2 = \frac{\mu_0 i_2}{2r}$,we get $B_{net} = \frac{\mu_0}{2r} \sqrt{i_1^2 + i_2^2}$.Given $r = 2\pi \, cm = 2\pi 	imes 10^{-2} \, m$,$i_1 = 3 \, A$,$i_2 = 4 \, A$,and $\mu_0 = 4\pi 	imes 10^{-7} \, Wb/A \cdot m$.$B_{net} = \frac{4\pi 	imes 10^{-7}}{2 	imes 2\pi 	imes 10^{-2}} \sqrt{3^2 + 4^2}$.$B_{net} = \frac{4\pi 	imes 10^{-7}}{4\pi 	imes 10^{-2}} \sqrt{9 + 16} = 10^{-5} 	imes \sqrt{25} = 5 	imes 10^{-5} \, Wb/m^2$.

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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