Ch 28 Magnetic Sources

Ch 28 Magnetic Sources - 2 sin 4 r qv B φ π μ = Note...

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Unformatted text preview: 2 sin 4 r qv B φ π μ = Note that each point P resides on a circle having constant magnitude B . As the angle φ decreases between r and v so too does B . μ The symbol: Is read mu-nought (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 ˆ 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 = ˆ 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 ˆ 4 r r x v q B π μ = r r q r r q E 3 2 4 1 ˆ 4 1 πε πε = = In full vector form, the B field, due to a moving point charge, is given by: 2 ˆ 4 r r x v q B π μ = q v P φ 2 sin 4 r qv B φ π μ = The SI unit for magnetic Field B is Tesla 1T = 1 N·s/C·m = 1 N/A·m It will be found when we study electromagnetic waves the speed of light in a vacuum is related to mu- nought and epsilon-nought: 2 1 ε μ = 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 much-much 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 ˆ 4 r r x v q B π μ = 2 ˆ 4 r r x v q B π μ = r r q E ˆ 4 1 2 πε = 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 ˆ 4 1 2 πε = And the repulsive force is simply E times the top charge of the proton: 2 ˆ 4 r r x v q B π μ = 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 ˆ 4 r r x v q B π μ = B x v q F B = The lower red proton creates a B field as it moves through spacetime with speed v. φ r v Note the B fields decreases as the charge passes by. The problem states the velocity is perpendicular to the ř and so provides the max angle for creating a B field....
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Ch 28 Magnetic Sources - 2 sin 4 r qv B φ π μ = Note...

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