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(D) The magnetic field $B$ at a distance $R$ from a long straight current-carrying wire is given by the formula:$B = \frac{\mu_0 I}{2\pi R}$Rearrangin

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$80$

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(D) The magnetic field $B$ at a distance $R$ from a long straight current-carrying wire is given by the formula:$B = \frac{\mu_0 I}{2\pi R}$Rearranging the formula to solve for $R$:$R = \frac{\mu_0 I}{2\pi B}$Given values:$I = 12 \ A$$B = 3 	imes 10^{-5} \ Wb/m^2$$\mu_0 = 4\pi 	imes 10^{-7} \ Wb/Am$Substituting the values into the equation:$R = \frac{4\pi 	imes 10^{-7} 	imes 12}{2\pi 	imes 3 	imes 10^{-5}}$$R = \frac{2 	imes 10^{-7} 	imes 12}{3 	imes 10^{-5}}$$R = \frac{24 	imes 10^{-7}}{3 	imes 10^{-5}}$$R = 8 	imes 10^{-2} \ m$To convert meters to millimeters $(mm)$,multiply by $1000$:$R = 8 	imes 10^{-2} 	imes 10^3 \ mm = 80 \ mm$

The magnetic field at a point $P$ situated at a perpendicular distance $R$ from a long straight wire carrying a current of $12 \ A$ is $3 \times 10^{-5} \ Wb/m^2$. The value of $R$ in $mm$ is $\left[\mu_0 = 4\pi \times 10^{-7} \ Wb/Am\right]$

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