A current-carrying wire oriented north to south an laid over top a compass defle

Question

A current-carrying wire oriented north to south an laid over top a compass deflects the compass 18 degrees to the east. What is the magnitude of the magnetic field made by the current? Assume the strength of the magnetic field due to the earth is about 2 times 10^5 [T]. Also, In what direction does the electron current flow in the wire (not conventional current, but electron current)? 6.5 times 10^-5 [T], the electron current flows from south to north 3.1 times 10^-5 [T], the electron current flows from south to north 9.3 times 10^-5 [T], the electron current flows from north to south 6.4 times 10^-5 [T], the electron current flows from north to south

Explanation / Answer

we would draw the compass needle, from the pivot point to its pointy tip, and have it pointing 18 deg east of north. Then the force on the tip is composed of two components: one from Earth's field, with magnitude 2*10^-5 T, and pulling North; and one from the wire's field, with unknown magnitude, and pulling East.
Once you've drawn that force diagram, note that tan(18 deg) = B/E, where B is the wire's B-field strength, and E is Earth's B-field strength.

direction from south to north....

Then B = 2*10^-5 T * tan(18 deg) = 6.498 * 10^-6

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