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

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$5 	imes 10^{-6} \, T$ Southward

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(D) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula:$B = \frac{\mu_0 I}{2 \pi r}$Given:$I = 100 \, A$$r = 4 \, m$$\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$Substituting the values:$B = \frac{(4\pi 	imes 10^{-7}) 	imes 100}{2 \pi 	imes 4}$$B = \frac{2 	imes 10^{-7} 	imes 100}{4}$$B = \frac{2 	imes 10^{-5}}{4} = 0.5 	imes 10^{-5} = 5 	imes 10^{-6} \, T$According to the Right-Hand Thumb Rule,if the thumb points in the direction of the current (West),the curled fingers at the point below the wire will point towards the South.

$A$ horizontal overhead powerline is at a height of $4\ m$ from the ground and carries a current of $100\ A$ from east to west. The magnetic field directly below it on the ground is: (Given: $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$)

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