Aconductor wire with length of 0.20 m and weight of 1.47 n is oriented perpendicularly to a magnetic field. if the current of 1.75a flows to the north, the net force on the wire is measured using a balance and found to be zero. what is the magnitude and direction of the magnetic field?

1. Magnitude: 4.4 T

The net force on the wire is zero, therefore the magnetic force must be equal to the weight of the wire:

Where the magnetic force is

where

I = 1.75 A is the current

B is the magnetic field

L = 0.20 m is the length of the wire

is the angle between the direction of L and B

Solving the formula for B, we find the magnitude of the magnetic field:

2. Direction: westward

The direction is given by the right's hand rule. We know that:

- thumb --> magnetic force (upward)

- index finger --> current (north)

- middle finger --> direction of magnetic field: so, it must be westward.

- middle

pe = weightxheight = 2000x1000=2,000,000joules