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(C) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 I}{2a}$,where $a$ is the radius.Given $a = 2\pi 	ext{ cm} = 2\pi

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$5 	imes 10^{-5}$

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(C) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 I}{2a}$,where $a$ is the radius.Given $a = 2\pi 	ext{ cm} = 2\pi 	imes 10^{-2} 	ext{ m}$.For the first coil with current $I_1 = 3 	ext{ A}$:$B_1 = \frac{\mu_0 	imes 3}{2 	imes 2\pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 3}{4\pi 	imes 10^{-2}} = 3 	imes 10^{-5} 	ext{ T}$.For the second coil with current $I_2 = 4 	ext{ A}$:$B_2 = \frac{\mu_0 	imes 4}{2 	imes 2\pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 4}{4\pi 	imes 10^{-2}} = 4 	imes 10^{-5} 	ext{ T}$.Since the coils are placed at right angles,the resultant magnetic field is $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{(3 	imes 10^{-5})^2 + (4 	imes 10^{-5})^2} = \sqrt{9 	imes 10^{-10} + 16 	imes 10^{-10}} = \sqrt{25 	imes 10^{-10}} = 5 	imes 10^{-5} 	ext{ T}$.

Two concentric coils each of radius equal to $2\pi \text{ cm}$ are placed at right angles to each other. If $3 \text{ A}$ and $4 \text{ A}$ are the currents flowing through the two coils respectively,the magnetic induction (in $\text{Wb m}^{-2}$) at the centre of the coils will be:

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