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Ch 22: Electromagnetic Induction. We’ve learned that electric currents generate magnetic fields. Now we will see how magnetism can generate electric currents. Motional EMF. Magnetic field exerts a force on moving conduction e- in the wire. Force is down b/c e- are negative.
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Ch 22: Electromagnetic Induction We’ve learned that electric currents generate magnetic fields. Now we will see how magnetism can generate electric currents.
Motional EMF Magnetic field exerts a force on moving conduction e- in the wire. Force is down b/c e- are negative
Electric Field is created e- are pushed to lower end Separation of charge creates electric field, E, within wire (pointing downward) A charge q in the wire feels two forces: FE=qE FB=qvBsinθ or qvB because θ=90o Since q is negative: FE is upward FB is downward Once FE = FB, the charges in wire are in electromagnetic equilibrium qE=qvB Or E=vB FORCES ARE IN OPPOSITE DIRECTIONS
Electric Potential Which end has the higher electric potential, point b or a? Answer: Point b The further b is from a the greater the electric potential
Potential Difference • Vba = El(hint: remember Vicky and Ed??) • Since E=vB Vba = vBl
Creating motional emf (electromotive force) • Rod sliding along conducting rails connected at left by a stationary bar • Sliding rod completes circuit • Potential difference Vba causes current to flow • Motion of sliding rod through magnetic field creates motional emf: ε=Blv
Existence of a current in sliding rod causes B to exert FB on it • FB=BIl (left b/c current is upward) • An external agent must provide same force to right to keep rod moving at constant velocity. • Power external agent must supply is • Electric Power delivered is P=IVba = Iε=I(Blv) • Note these two expressions are identical • Energy provided by external agent is transformed first into electrical energy and then thermal energy as conductors making up the circuit dissipate heat.
Magnetic Flux: A measure of the number of magnetic field lines that cross a given area (No field lines cross the surface) (only component of B that’s perpendicular to surface)
Magnetic Flux Units 1 Tm2 1 Weber = 1 Wb
The figure to the right shows two views of a circular loop of radius 3 cm placed within a uniform magnetic field, B=.2TWhat’s the magnetic flux through the loop? Since, B is parallel to A
What would be the magnetic flux through the loop if the loop were rotated 45 degrees?
What would be the magnetic flux through the loop if the loop were rotated 90o? If the angle between B and A is 90 degrees, magnetic flux through the loop is zero, since cos 90o = 0
Faraday’s Law of Electromagnetic Induction • Changes in magnetic flux induce emf • Magnitude of emf induced equals rate of change of magnetic flux in a circuit
Lenz’s Law • Induced emf can produce a current • Direction of current determined by polarity of induced emf • Current will always flow in the direction opposing the change in magnetic flux that produced it. • Otherwise, energy would not be conserved
The circular loop from our previous problem rotates at a constant angular speed through 45 degrees in 0.5s.What’s the induced emf in the loop?
In which direction will current be induced to flow? Φ=.00057Wb upward Φ=.00040Wb upward Change in magnetic flux is -.00017Wb upward This equals +.00017Wb downward To oppose this change, we need some magnetic field upward If current flows counter-clockwise, a magnetic field is created pointing upward inside the ring
AREA INCREASES Again consider the conducting rod that’s moving with constant velocity, v, along a pair of parallel conducting rails (separated by a distance l ), within a uniform magnetic field, B. Find the induced emf and the direction of the induced current in the rectangular circuit. As rod moves to the right, magnetic flux into page increases Therefore, a magnetic field out of the page is needed to oppose this Therefore, the current must travel counterclockwise
Φ is downward and increases as magnet moves. A permanent magnet creates a magnetic field in the surrounding space. The end of the magnet at which the field lines emerge is designated the north pole (N) and the other end the south pole (S). The figure below shows a bar magnet moving down, through a circular loop of wire. What will be the direction of the induced current in the wire? ANSWER: COUNTERCLOCKWISE to create an upward flux According to Lenz’s Law: the induced emf will generate a current that opposes this change
What will be the direction of the induced current in the wire if the magnet is moved as shown in the following diagram? Magnetic flux is UPWARD Therefore: Current is generated CLOCKWISE To create a magnetic flux DOWNWARD
A square loop of wire 2 cm on each side contains 5 tight turns and has a total resistance of .0002Ω. It is placed 20cm from a long, straight, current-carrying wire. If the current in the straight wire is increased at a steady rate from 20A to 50A in 2s, determine the magnitude and direction of the current induced in the square loop. (Because the square loop is at such a great distance from the straight wire, assume that the magneitc field through the loop is uniform and equal to the magnetic field at its center.) Calculate the emf induced in the loop: Calculate value of current: Since magnetic flux from straight wire is OUT OF PAGE Induced current is CLOCKWISE to create an opposing flux INTO THE PAGE