Chapter 9 - Magnetic Forces, Materials and Inductance

Chapter 9 - Magnetic Forces, Materials and Inductance -...

Info iconThis preview shows pages 1–14. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 9 Magnetic Forces, Materials, and Inductance Magnetic Forces, Materials, and Inductance Physical significance of Chapter 8 Magnetic forces and torques Magnetic materials Inductance
Background image of page 1

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

View Full DocumentRight Arrow Icon
In electrostatics, E causes a force to be exerted on a charge (either stationary or moving) In steady magnetic fields, H causes a force to be exerted ONLY on moving charges. Force on a Moving Charge
Background image of page 2
Force on a Moving Charge In electrostatics, F = Q E Given a charged particle in motion in a magnetic field of flux density B , Force exerted on this particle F = Q v x B Where v = velocity of charge Note: The acceleration vector is always normal to v .
Background image of page 3

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

View Full DocumentRight Arrow Icon
Trajectory of a moving electron in a region with magnetic field.
Background image of page 4
Thus, the magnitude of velocity is unchanged. There is no change in kinetic energy. Therefore, H is incapable of transferring energy to the moving charge. If both E and B are present, F = Q ( E + ( v x B )) Lorentz Force Equation
Background image of page 5

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

View Full DocumentRight Arrow Icon
Example: Cyclotron The cyclotron uses electric and magnetic fields to accelerate charged particles
Background image of page 6
d F = dQ v x B d F = ρ v dv v x B Force on a Differential Current Element • the force due to H is exerted on moving charges confined in the conductor • in effect, this force is collectively transferred to the conductor itself Since J = ρ v v , d F = J x B dv
Background image of page 7

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

View Full DocumentRight Arrow Icon
d F = J x B dv J dv = K dS = I dL dF = K x B dS dF = I d L x B × - = × = L B B L F d I Id For a closed circuit: Over the entire volume or surface: dv vol B J F × = × = s dS B K F
Background image of page 8
For a straight conductor in a uniform magnetic field: F = I L x B I F I = F = B I L sin θ dF = I d L x B
Background image of page 9

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

View Full DocumentRight Arrow Icon
Consider a square loop of wire and a current-carrying filament H due to filament at the xy-plane: m A x 2 15 x 2 I z z a a π = π = H H B 7 0 10 4 - × π = μ = T x 10 3 6 z a - × =
Background image of page 10
What is F on the loop? × - = L B F d I loop pN 8 ) dy ( 1 ) dx ( x ) dy ( 3 ) dx ( x 10 3 mA 2 2 0 y 3 1 x 2 0 y 3 1 x 6 x y z x z y z x z a a a a a a a a a F - = - × + - × + × + × - = = = = = - T x 10 3 6 z a B - × =
Background image of page 11

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

View Full DocumentRight Arrow Icon
2 2 2 2 2 2 2 2 2 12 1 1 2 d d I ) d ( d d I d R 4 d I d B L F B L F a L H r12 × = × = π × = I 2 d L 2 I 1 d L 1 R 1 2 Force Between Differential Current Elements
Background image of page 12
2 12 1 1 0 2 2 R 4 x d I d I π μ × = 12 R a L L ( ) [ ] 12 R a L L x
Background image of page 13

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

View Full DocumentRight Arrow Icon
Image of page 14
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 49

Chapter 9 - Magnetic Forces, Materials and Inductance -...

This preview shows document pages 1 - 14. Sign up to view the full document.

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