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(A) 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 v

Vedclass · Accepted Answer

$1.2 	imes 10^{-5} 	ext{ T}$,towards north

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(A) 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 values are current $I = 90 	ext{ A}$ and distance $r = 1.5 	ext{ m}$.The permeability of free space is $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.Substituting these values into the formula:$B = \frac{4\pi 	imes 10^{-7} 	imes 90}{2\pi 	imes 1.5}$$B = \frac{2 	imes 10^{-7} 	imes 90}{1.5}$$B = 2 	imes 10^{-7} 	imes 60 = 120 	imes 10^{-7} = 1.2 	imes 10^{-5} 	ext{ T}$.According to the right-hand thumb rule,if the current flows from east to west,the magnetic field at a point above the wire points towards the north.

$A$ horizontal overhead power line carries a current of $90 \text{ A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 \text{ m}$ above the line?

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