The magnetic field at the centre O of a wire | vedclass

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(A) The magnetic field at the centre of a semicircular arc of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4R}$.For the two semicir

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$3$

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(A) The magnetic field at the centre of a semicircular arc of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4R}$.For the two semicircular arcs of radii $R_1$ and $R_2$,the magnetic fields at the centre $O$ are in the same direction (inward,perpendicular to the plane of the loop).$B_{total} = B_1 + B_2 = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2} = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} + \frac{1}{R_2} ight)$.Given $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$,$I = 4 	ext{ A}$,$R_1 = 2\pi 	ext{ m}$,and $R_2 = 4\pi 	ext{ m}$.$B_{total} = \frac{4\pi 	imes 10^{-7} 	imes 4}{4} \left( \frac{1}{2\pi} + \frac{1}{4\pi} ight) = 4\pi 	imes 10^{-7} \left( \frac{2+1}{4\pi} ight) = 4\pi 	imes 10^{-7} 	imes \frac{3}{4\pi} = 3 	imes 10^{-7} 	ext{ T}$.Comparing this with $\alpha 	imes 10^{-7} 	ext{ T}$,we get $\alpha = 3$.

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