A straight 2.70 m wire carries a typical household current of 1.50 a (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.

a) F = 2.2275 * 10^-4 N , upwards

b) F = 2.2275 * 10^-4 N , east to west

c) F = 0

Explanation:

Given:

- The straight wire of length  L = 2.7 m

- The current I in the wire I = 1.50 A

- The magnetic Field B = 0.55 * 10^-4 T

Find:

Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running

a) from west to east

b) vertically upward

c) from north to south

d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

Solution:

- If current runs from west to east the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(90)*10^-4

                           F = 2.2275 * 10^-4 N

- From Figure B points north and current I points east. From right hand rule, the direction of force is out of page, so its upward.

- If current runs upward the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(90)*10^-4

                           F = 2.2275 * 10^-4 N

- From Figure B points north and current I points out of page . From right hand rule, the direction of force is out of page, so its east to west.

- If current runs north to south the angle between the magnetic field B is θ = 0°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(0)*10^-4

                           F = 0 N

need to see circuit to get time constant