Question

A straight 2.20 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. Part A Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east. SubmitRequest Answer Part B Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east. F1 F 1 = nothing N SubmitRequest Answer Part C

Answer 1

A) Upward

In order to find the direction of the magnetic force on the wire, we can use the right-hand rule: the index finger, the middle finger and the thumb of the right hand must be placed all of them perpendicular to each other.

So we have:

- Index finger: direction of current in the wire (from west to east)

- Middle finger: direction of magnetic field (from south to north)

- Thumb: direction of the force --> so it will be upward

So, the force will point upward.

B)
`1.82\cdot 10^(-4)N`

The magnitude of the force exerted by the magnetic field on the wire is given by

`F=ILB`

where

I = 1.50 A is the current in the wire

L = 2.20 m is the length of the wire

`B=0.550 G = 0.55 \cdot 10^(-4)T`

Substituting into the equation, we find

`F=(1.50 A)(2.20 m)(0.55 \cdot 10^(-4) T)=1.82\cdot 10^(-4)N`