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(B) The magnetic field $B$ at the centre of a circular loop carrying current $i$ is given by $B_{	ext{loop}} = \frac{\mu_0 i}{2r}$ (outwards).The mag

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$3.4 	imes 10^{-5} \,T$,outwards

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(B) The magnetic field $B$ at the centre of a circular loop carrying current $i$ is given by $B_{	ext{loop}} = \frac{\mu_0 i}{2r}$ (outwards).The magnetic field $B$ due to a long straight wire at a distance $r$ is $B_{	ext{wire}} = \frac{\mu_0 i}{2\pi r}$ (inwards).Given: $i = 8\,A$,$r = 10\,cm = 0.1\,m$.The net magnetic field at the centre $C$ is $B_C = B_{	ext{loop}} - B_{	ext{wire}}$.$B_C = \frac{\mu_0 i}{2r} - \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{2r} \left(1 - \frac{1}{\pi}ight)$.Substituting the values:$B_C = \frac{4\pi 	imes 10^{-7} 	imes 8}{2 	imes 0.1} \left(1 - \frac{1}{3.14}ight)$.$B_C = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 8}{0.2} 	imes \left(\frac{2.14}{3.14}ight)$.$B_C = 160 	imes 10^{-7} 	imes 2.14 = 342.4 	imes 10^{-7} = 3.424 	imes 10^{-5} \,T$.Since $B_{	ext{loop}} > B_{	ext{wire}}$,the direction is outwards.

$A$ long,straight wire is turned into a loop of radius $10\,cm$ (see figure). If a current of $8\,A$ is passed through the loop,then the value of the magnetic field and its direction at the centre $C$ of the loop shall be close to

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