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(B) 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}{4\pi} \frac{2I}{r

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

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(B) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula: $B = \frac{\mu_0}{4\pi} \frac{2I}{r}$ Given: Current $I = 100\, A$ Distance $r = 4\, m$ Permeability constant $\frac{\mu_0}{4\pi} = 10^{-7}\, T\, m A^{-1}$ Substituting the values: $B = 10^{-7} 	imes \frac{2 	imes 100}{4} = 10^{-7} 	imes 50 = 5 	imes 10^{-6}\, T$ According to the Right-Hand Thumb Rule,if the current flows from west to east,the magnetic field lines circle the wire. Directly below the wire,the magnetic field is directed 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 west to east. The magnetic field directly below it on the ground is $(\mu_0 = 4\pi \times 10^{-7}\, T\, m A^{-1})$.

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