A long straight wire in the horizontal plane | vedclass

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(D) Current in the wire,$I = 50 \; A$.The point is $2.5 \; m$ away to the east of the wire.Therefore,the distance of the point from the wire is $r = 2

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$4 	imes 10^{-6} \; T$

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(D) Current in the wire,$I = 50 \; A$.The point is $2.5 \; m$ away to the east of the wire.Therefore,the distance of the point from the wire is $r = 2.5 \; m$.The magnitude of the magnetic field $B$ at a distance $r$ from a long straight wire is given by the formula:$B = \frac{\mu_0 I}{2 \pi r} = \frac{\mu_0 (2I)}{4 \pi r}$Substituting the values:$B = \frac{4 \pi 	imes 10^{-7} 	imes 50}{2 \pi 	imes 2.5}$$B = \frac{2 	imes 10^{-7} 	imes 50}{2.5}$$B = \frac{100 	imes 10^{-7}}{2.5} = 40 	imes 10^{-7} = 4 	imes 10^{-6} \; T$.According to the Right-Hand Thumb Rule,if the current flows from north to south,the magnetic field lines circle the wire. At a point to the east of the wire,the direction of the magnetic field is vertically upward.

$A$ long straight wire in the horizontal plane carries a current of $50 \; A$ in the north to south direction. Give the magnitude and direction of $B$ at a point $2.5 \; m$ east of the wire.

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