A horizontal conductor is oriented north-sout | vedclass

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(B) $1$. The magnetic field $B$ produced by a current-carrying conductor oriented north-south can be determined using the Right-Hand Thumb Rule. For a

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South

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(B) $1$. The magnetic field $B$ produced by a current-carrying conductor oriented north-south can be determined using the Right-Hand Thumb Rule. For a conductor carrying current from south to north,the magnetic field above the conductor points towards the west.$2$. The force on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} 	imes \vec{B})$.$3$. Given the particle is positively charged $(q > 0)$,moves north ($\vec{v}$ is north),and experiences an upward force ($\vec{F}$ is up),we confirm the magnetic field $\vec{B}$ at that position points west.$4$. Now,consider the particle located to the east of the conductor. The magnetic field $\vec{B}$ produced by the conductor at a point to its east points vertically upward.$5$. The particle has a velocity $\vec{v}$ directed towards the conductor (i.e.,towards the west).$6$. Applying the right-hand rule for the cross product $\vec{v} 	imes \vec{B}$: $\vec{v}$ is west and $\vec{B}$ is up. The resulting force $\vec{F}$ points towards the south.

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