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

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

Step-by-step 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}* \vec{B})`

`|\vec{F}| = I |(\vec{L}* \vec{B})|`

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

`|\vec{F}| = 5500 * 160 * 60 * 10^(-6)* 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.