A magnetic field of 5 times 10^{-5} ,T is pro | vedclass

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(C) Given: Magnetic field $B = 5 	imes 10^{-5} \,T$, distance $r = 0.2 \,m$, and permeability of free space $\mu_0 = 4 \pi 	imes 10^{-7} \,T \cdot m

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$50$

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(C) Given: Magnetic field $B = 5 	imes 10^{-5} \,T$, distance $r = 0.2 \,m$, and permeability of free space $\mu_0 = 4 \pi 	imes 10^{-7} \,T \cdot m/A$.The magnetic field produced by a long straight current-carrying wire is given by the formula:$B = \frac{\mu_0}{2 \pi} \cdot \frac{I}{r}$Substituting the given values into the equation:$5 	imes 10^{-5} = \frac{4 \pi 	imes 10^{-7}}{2 \pi} 	imes \frac{I}{0.2}$$5 	imes 10^{-5} = 2 	imes 10^{-7} 	imes \frac{I}{0.2}$$5 	imes 10^{-5} = 10^{-6} 	imes I$$I = \frac{5 	imes 10^{-5}}{10^{-6}} = 5 	imes 10^1 = 50 \,A$Therefore, the current passing through the wire is $50 \,A$.

$A$ magnetic field of $5 \times 10^{-5} \,T$ is produced at a perpendicular distance of $0.2 \,m$ from a long straight wire carrying electric current. If the permeability of free space is $4 \pi \times 10^{-7} \,T \cdot m/A$, the current passing through the wire in $A$ is:

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