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(D) The magnetic field $B$ at the center of a circular coil of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.Here,$r = 20\,cm =

Vedclass · Accepted Answer

$628$

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(D) The magnetic field $B$ at the center of a circular coil of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.Here,$r = 20\,cm = 0.2\,m$ and $i = \sqrt{2}\,A$.Since the two coils are identical and placed in perpendicular planes,the magnetic field produced by each coil at the center will have the same magnitude $B_C$ but will be directed along perpendicular axes (e.g.,$x$ and $y$ axes).$B_C = \frac{4\pi 	imes 10^{-7} 	imes \sqrt{2}}{2 	imes 0.2} = \pi 	imes \sqrt{2} 	imes 10^{-6}\,T$.The net magnetic field $B_{net}$ at the center is the vector sum of the two fields:$B_{net} = \sqrt{B_C^2 + B_C^2} = B_C \sqrt{2}$.Substituting the value of $B_C$:$B_{net} = (\pi 	imes \sqrt{2} 	imes 10^{-6}) 	imes \sqrt{2} = 2\pi 	imes 10^{-6}\,T$.$B_{net} = 2 	imes 3.14 	imes 10^{-6}\,T = 6.28 	imes 10^{-6}\,T$.$B_{net} = 628 	imes 10^{-8}\,T$.Thus,the value is $628$.

Two identical circular wires of radius $20\,cm$ and carrying current $\sqrt{2}\,A$ are placed in perpendicular planes as shown in the figure. The net magnetic field at the centre of the circular wires is $.............\times 10^{-8}\,T$. (Take $\pi=3.14$ )

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