This preview shows pages 1–10. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 2 sin 4 r qv B = Note that each point P resides on a circle having constant magnitude B . As the angle decreases between r and v so too does B . The symbol: Is read munought (without the subscript 0 it is the Greek letter mu. What does sine function lead us to think? Magnitude of B To avoid saying the direction from the source q to the field point P over and over again, we introduce the unit vector: r This vector is also called r hat that points from the source point to the field point P . This unit vector is equal to the position vector r divided by its magnitude. r r r = A point charge in motion also produces an electric field, with field lines that radiate outward from a positive charge. The magnetic field lines are completely different. As we have seen magnetic field lines are always circles or loops that close on themselves; not radiating lines as with electric field lines. 2 4 r r x v q B = r r q r r q E 3 2 4 1 4 1 = = In full vector form, the B field, due to a moving point charge, is given by: 2 4 r r x v q B = q v P 2 sin 4 r qv B = The SI unit for magnetic Field B is Tesla 1T = 1 Ns/Cm = 1 N/Am It will be found when we study electromagnetic waves the speed of light in a vacuum is related to mu nought and epsilonnought: 2 1 = c Example of Forces between Two Moving Protons Two protons move at the same speed v parallel but in the opposite directions . Their speed is muchmuch smaller than c . This is necessary or we would have to consider special relativistic effects. At the instant shown find: 1) Electric and magnetic forces on the upper proton 2) Determine the ratio of their magnitudes. 2 4 r r x v q B = 2 4 r r x v q B = r r q E 4 1 2 = As long as the velocity is much smaller than c , the electric field arrives at the top proton well before the bottom proton moves significantly. Hence, an electric field is present do to q proton at the bottom acting on q proton at the top. The electric force field is given by: r r q q F p p E 4 1 2 = And the repulsive force is simply E times the top charge of the proton: 2 4 r r x v q B = Now the magnetic force for a charge moving through a magnetic field at velocity v is given: B x v q F B = However for a point charge, the magnetic field was experimentally determined to be: 2 4 r r x v q B = B x v q F B = The lower red proton creates a B field as it moves through spacetime with speed v. r v Note the B fields decreases as the charge passes by. The problem states the velocity is perpendicular to the and so provides the max angle for creating a B field....
View
Full
Document
This note was uploaded on 10/05/2011 for the course PHYS 4B taught by Professor W.christensen during the Summer '09 term at Irvine Valley College.
 Summer '09
 W.CHRISTENSEN
 Physics

Click to edit the document details