Two circular coils P and Q of 100 turns each | vedclass

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(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.Given: $N = 100$,$r = \pi 	ext{ cm} = \pi 	imes 10^{-

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$20$

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(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.Given: $N = 100$,$r = \pi 	ext{ cm} = \pi 	imes 10^{-2} 	ext{ m}$.For coil $P$ with $I_1 = 1 	ext{ A}$:$B_P = \frac{\mu_0 	imes 100 	imes 1}{2 	imes \pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 100}{2 \pi 	imes 10^{-2}} = 2 	imes 10^{-3} 	ext{ T} = 2 	ext{ mT}$.For coil $Q$ with $I_2 = 2 	ext{ A}$:$B_Q = \frac{\mu_0 	imes 100 	imes 2}{2 	imes \pi 	imes 10^{-2}} = \frac{4\pi 	imes 10^{-7} 	imes 200}{2 \pi 	imes 10^{-2}} = 4 	imes 10^{-3} 	ext{ T} = 4 	ext{ mT}$.Since the planes are mutually perpendicular,the magnetic fields $B_P$ and $B_Q$ are perpendicular to each other.The resultant magnetic field is $B_{	ext{net}} = \sqrt{B_P^2 + B_Q^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} 	ext{ mT}$.Comparing with $\sqrt{x} 	ext{ mT}$,we get $x = 20$.

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