Recent content by gtqueen371

I didn't even think about gravity...I was thinking electrical and magnetic forces. :( THANKS!

mg = I\DeltaL x B (.08kg)(9.81 m/s/s) = (3A)(.14m) x B 1.869 T = B in the -z direction! Thanks for your help!

gravitational and magnetic

The net force must equal zero. So F = qE + qvB becomes -qE = qvB which then is E=vB but I don't have E or v. I don't understand how this fits together with Fmagnetic = I\DeltaL x B... Also if I use Fmagnetic = I\DeltaL x B with -qE = qVB, then I get -qE = I\DeltaL x B which then equals...

Homework Statement A metal rod of length 14 cm and mass 80 grams has metal loops at both ends, which go around two metal poles. The rod is in good electrical contact with the poles but can slide freely up and down. A magnet supplies a large uniform magnetic field B in the region of the rod...

Three bulbs A circuit is made of two 1.5 volt batteries and three light bulbs as shown in Figure 19.57. When the switch is closed and the bulbs are glowing, bulb 1 has a resistance of 9 ohms, bulb 2 has a resistance of 44 ohms, bulb 3 has a resistance of 28 ohms, and the copper connecting wires...