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(B) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by the formula:$B = \frac{\mu_0}{4\pi} \frac{2

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

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(B) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by the formula:$B = \frac{\mu_0}{4\pi} \frac{2I}{r}$Given:$I = 100\,A$$r = 4\,m$$\frac{\mu_0}{4\pi} = 10^{-7}\,TmA^{-1}$Substituting the values:$B = 10^{-7} 	imes \frac{2 	imes 100}{4}$$B = 10^{-7} 	imes 50 = 5 	imes 10^{-6}\,T$According to the Right-Hand Thumb Rule,if the current flows from east to west,the magnetic field lines form concentric circles around the wire. Directly below the wire,the direction of the magnetic field is towards the south.

$A$ horizontal overhead power line 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 $(\mu _0 = 4\pi \times 10^{-7}\,TmA^{-1})$

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