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(A) The magnetic field at the center of a circular coil of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.Given $r = 4 \pi 	ext

Vedclass · Accepted Answer

$13 	imes 10^{-5} 	ext{ Wb m}^{-2}$

Vedclass · Answer

(A) The magnetic field at the center of a circular coil of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.Given $r = 4 \pi 	ext{ cm} = 4 \pi 	imes 10^{-2} 	ext{ m}$.For the first coil with current $i_1 = 10 	ext{ A}$:$B_1 = \frac{\mu_0 	imes 10}{2 	imes 4 \pi 	imes 10^{-2}} = \frac{4 \pi 	imes 10^{-7} 	imes 10}{8 \pi 	imes 10^{-2}} = 0.5 	imes 10^{-4} 	ext{ T} = 5 	imes 10^{-5} 	ext{ T}$.For the second coil with current $i_2 = 24 	ext{ A}$:$B_2 = \frac{\mu_0 	imes 24}{2 	imes 4 \pi 	imes 10^{-2}} = \frac{4 \pi 	imes 10^{-7} 	imes 24}{8 \pi 	imes 10^{-2}} = 1.2 	imes 10^{-4} 	ext{ T} = 12 	imes 10^{-5} 	ext{ T}$.Since the coils are at right angles,the resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{(5 	imes 10^{-5})^2 + (12 	imes 10^{-5})^2} = \sqrt{25 + 144} 	imes 10^{-5} 	ext{ T} = \sqrt{169} 	imes 10^{-5} 	ext{ T} = 13 	imes 10^{-5} 	ext{ Wb m}^{-2}$.

Two concentric coils each of radius equal to $4 \pi \text{ cm}$ are placed at right angles to each other. If $10 \text{ A}$ and $24 \text{ A}$ are the currents flowing through the coils,respectively,the magnetic induction at the center of the coils will be

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