Bill and Bev

Force on a Moving Charge B The force on a charge q, is F = Bqv where v is the velocity q F v

The force causes the particle to travel in a circular path Link to Circular Paths The force causes the particle to travel in a circular path

Link to Circular Paths F = Bqv Bqv = mv2 F = mv2 r r r = mv Bq The radius of curvature depends on the strength of the field, the mass of the particle, the charge of the particle and the velocity.

Mass Spectrometer

Link to Circular Paths T = 2pir v T = 2 pi m BQ r = mv Bq

Questions A particle of mass 9.1 x 10-31 kg and with a charge of 1.6 x 10-19 C, is moving at 4.5 x 106 ms-1. It enters a uniform magnetic field of flux density 0.15 mT at 90 degrees. What is the radius of its circular path.

Magnetic Flux The magnetic flux density (B) x the area swept out (A) = magnetic flux (theta) Units Webber Dt bring

Electromagnetic Induction A moving charge in a magnetic field experiences a force (to make it move). Therefore moving a conductor in a field will cause a current to flow. This is electromagnetic induction. Or a varying magnetic field over a conductor will also cause a current.

Electromagnetic Induction Flux cutting. The conductor has to cut field lines for an emf to be induced. Discussion: How is the ‘electricity’ made? The demonstrations have shown that ‘making’ electricity involves magnetic fields, but what is really going on? Your students already know that charges moving across a magnetic field experience a force (the BIL force). Now, the metal of a conductor contains mobile charges, the conduction electrons. What happens to these if the conductor is moved across a magnetic field? Consider a conducting rod PQ moving at a steady speed v perpendicular to a field with a flux density B. An electron (negative charge e) in the rod will experience a force (= Bev) (Fleming's left hand rule) that will push it towards the end Q. The same is true for other electrons in the rod, so the end Q will become negatively charged, leaving P with a positive charge. As a result, an electric field E builds up until the force on electrons in the rod due to this electric field (= Ee) balances the force due to the magnetic field. Ee = Bev so E =Bv For a rod of length L, E = V/L and so V/L = Bv Hence the induced EMF E = BLv Clearly what we have here is an induced EMF (no complete circuit so no current flows) and already we can see that more rapid movement gives a greater induced EMF. Now consider what happens when the EMF drives a current in an external circuit. To do this, imagine that the rod moves along a pair of parallel conductors that are connected to an external circuit.

Magnetic Flux Linkage Through a coil of N turns n theta = n BA N theta = NBAcos theta when the magnetic field is along the normal (perpendicular) of the coil face the N theapta = NBA When the coil is turned 180 then N theta = -NBA When the magnetic field is parallel to the coil flux linkage = 0

Faraday's Law The induced emf in a conductor is equal to the rate of change of flux linkage through the circuit. Write this out as an equation. Derive the equation from V= W/Q