hw07 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 7. 28.35, 29.30, 29.36, 29.87; Q30.02 28.35. a) The magnetic force on the wire must produce an upward force equal to the weight of the wire if the tension in the leads is to equal zero. The magnetic force is described by the equation, B L i F B = . Since L and B are perpendicular to each other, the magnitude of the magnetic force is given by, F B = iLB . Then, we must have, iLB = mg , so T) m)(0.440 620 . ( ) m/s kg)(9.8 0130 . ( 2 = = LB mg i i = 0.467 A b) B L i F B = . According to the right hand rule for the vector product, the current must be directed from left to right to produce an upward force with the magnetic field given. From left to right 29.30. Since all the currents are parallel, the forces between each pair of wires is attractive. From the symmetry, it should be clear that the magnitude of the net force on each wire is the same. Number the wires, starting with the lower left corner and proceeding clockwise. the same....
View Full Document

This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

Page1 / 3

hw07 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online