A straight wire carrying a current 10, A is b | vedclass

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(C) The magnetic field $B$ at the center of a semicircular arc carrying current $i$ and having radius $r$ is given by the formula:$B = \frac{\mu_0 i}{

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$6.28 	imes 10^{-5}\, T$

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(C) The magnetic field $B$ at the center of a semicircular arc carrying current $i$ and having radius $r$ is given by the formula:$B = \frac{\mu_0 i}{4r}$Given values are:$i = 10\, A$$r = 5\, cm = 5 	imes 10^{-2}\, m$$\mu_0 = 4\pi 	imes 10^{-7}\, T\cdot m/A$Substituting these values into the formula:$B = \frac{4\pi 	imes 10^{-7} 	imes 10}{4 	imes 5 	imes 10^{-2}}$$B = \frac{\pi 	imes 10^{-6}}{5 	imes 10^{-2}}$$B = \frac{3.14159}{5} 	imes 10^{-4} = 0.6283 	imes 10^{-4} = 6.28 	imes 10^{-5}\, T$Thus, the correct option is $C$.

$A$ straight wire carrying a current $10\, A$ is bent into a semicircular arc of radius $5\, cm.$ The magnitude of the magnetic field at the center is

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