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(B) The magnetic field $B$ at the surface of a wire carrying current $I$ is given by the formula derived from Ampere's Law: $B = \frac{\mu_0 I}{2 \pi

Vedclass · Accepted Answer

$4$

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(B) The magnetic field $B$ at the surface of a wire carrying current $I$ is given by the formula derived from Ampere's Law: $B = \frac{\mu_0 I}{2 \pi r}$.Here,the current $I = 12 \ A$.The diameter of the wire is $d = 1.2 \ mm = 1.2 	imes 10^{-3} \ m$.The radius $r = \frac{d}{2} = 0.6 	imes 10^{-3} \ m$.The permeability of free space $\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$.Substituting the values:$B = \frac{(4 \pi 	imes 10^{-7}) 	imes 12}{2 \pi 	imes 0.6 	imes 10^{-3}}$$B = \frac{2 	imes 10^{-7} 	imes 12}{0.6 	imes 10^{-3}}$$B = \frac{24 	imes 10^{-7}}{0.6 	imes 10^{-3}} = 40 	imes 10^{-4} \ T = 4 	imes 10^{-3} \ T = 4 \ mT$.Thus,the maximum magnetic field is $4 \ mT$.

The maximum magnetic field produced by a current of $12 \ A$ passing through a copper wire of diameter $1.2 \ mm$ is (in $mT$)

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