Question

A horizontal conductor in a power line carries a current of 5500 A from south to north. Earth's magnetic field (60.0 μT) is directed toward the north and is inclined downward at 67.0° to the horizontal. Find the magnitude and direction of the magnetic force on 160 m of the conductor due to Earth's field.

Answer 1

Answer:

$|\vec{F}| =48.60\ N$

Explanation:

given,

Current in the power line = I = 5500 A

earth's magnetic field = 60.0 μT

inclination downward = 67°

Length = 160 m

magnetic force = ?

$\vec{F} = I (\vec{L}\times \vec{B})$

$|\vec{F}| = I |(\vec{L}\times \vec{B})|$

$|\vec{F}| =I LB sin \theta$

$|\vec{F}| = 5500 \times 160 \times 60 \times 10^{-6}\times sin 67^0$

$|\vec{F}| =48.60\ N$

According to the right hand rule the direction of the force is perpendicular to the plane of the length and the magnetic field so, it is to west.