Five very long,straight wires are bound toget | vedclass

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(D) The net current $I_{net}$ flowing through the cable is the algebraic sum of the individual currents:$I_{net} = I_1 + I_2 + I_3 + I_4 + I_5 = 20 -

Vedclass · Accepted Answer

$74\,\mu T$

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(D) The net current $I_{net}$ flowing through the cable is the algebraic sum of the individual currents:$I_{net} = I_1 + I_2 + I_3 + I_4 + I_5 = 20 - 6 + 12 - 7 + 18 = 37\,A$Since the wires are bound together to form a small cable,we can treat the cable as a single long straight wire carrying the net current $I_{net}$ at a distance $r = 10\,cm = 0.1\,m$.The magnetic field $B$ due to a long straight wire is given by:$B = \frac{\mu_0}{4\pi} \cdot \frac{2I_{net}}{r}$Substituting the values:$B = 10^{-7} 	imes \frac{2 	imes 37}{0.1} = 10^{-7} 	imes 740 = 74 	imes 10^{-6}\,T = 74\,\mu T$Thus,the correct option is $D$.

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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