Physics Lecture 3 Magnetism

6
Physics Lecture 3 Magnetism: 1. Extended Lecture Notes- all notes so far and extra concept questions- DO FOR MID SEMESTER EXAM Concept Clicker Q’s: Force= qv x B Therefore, the velocity points to the right (thumb), magnetic field into the board, palm is facing up (force) C= Up -All positive charge carriers move to the top of the tube therefore, cannot be zero

description

:)

Transcript of Physics Lecture 3 Magnetism

Physics Lecture 3 Magnetism:

1. Extended Lecture Notes- all notes so far and extra concept questions- DO FOR MID SEMESTER EXAM

Concept Clicker Qs:

Force= qv x BTherefore, the velocity points to the right (thumb), magnetic field into the board, palm is facing up (force) C= Up

-All positive charge carriers move to the top of the tube therefore, cannot be zero-Therefore, a higher voltage on top, charge carriers moving upward Therefore A.

B= qv x B charge carrier has a negative charge, pointing upwards excess negative charge at the top- negative voltage

RHR: V x B, since q is of opposite direction, velocity to the left, field going into the board (finger) C points down, due to the negative q the field changes direction and goes out of the page.

Are we dealing with positrons or negatrons?-magnetic field out of the page, velocity is pointing up (yellow line) , force is curving around towards the right if q was positive, charge must be negative since it is going towards the left. - force is pushing the velocity to the side therefore ve. B

Which is positive/negative ?

Green= negative electrons- downwards Magnetic field going into the board Initial velocity is equal between positron and negatron Red= upwards

Motion in a uniform B- field F=maqvB=mv^2/rr=mv/qB v is constant F points toward the center of the circle

The centripetal acceleration points toward the center of the circle. It changes the particles direction but not the speed. -path of the beam becomes a helix

due to symmetry they cancel out = E= 0

D= down Symmetry- left and right cancel Top no magnetic fieldMagnitude of the net Force?? -A= iLB

Torque :