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(D) According to Ampere's circuital law,$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 \sum I_{enclosed}$.For a point at distance $r$ from the cable,the mag

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$74\ \mu T$

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(D) According to Ampere's circuital law,$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 \sum I_{enclosed}$.For a point at distance $r$ from the cable,the magnetic field $B$ is given by $B(2\pi r) = \mu_0 I_{net}$.The net current $I_{net} = I_1 + I_2 + I_3 + I_4 + I_5 = 20 - 6 + 12 - 7 + 18 = 37\ A$.Given $r = 10\ cm = 0.1\ m$ and $\mu_0 = 4\pi 	imes 10^{-7}\ T\cdot m/A$.Substituting the values: $B = \frac{\mu_0 I_{net}}{2\pi r} = \frac{4\pi 	imes 10^{-7} 	imes 37}{2\pi 	imes 0.1}$.$B = 2 	imes 10^{-7} 	imes 370 = 74 	imes 10^{-6}\ T = 74\ \mu T$.

Five very long,straight wires are bound together to form a small cable. Currents carried by the wires are $I_1 = 20\ A, I_2 = -6\ A, I_3 = 12\ A, I_4 = -7\ A, I_5 = 18\ A$. The magnetic induction at a distance of $10\ cm$ from the cable is

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