MAGNETIC INDUCTION

MAGNETIC INDUCTION MAGNETUIC FLUX: FARADAY’S LAW, INDUCED EMF: MOTIONAL EMF: THE AC GENERATOR: THE TRANSFORMER: INDUCTANCE: STORED ENERGY: Written by Dr. John Dayton

f = q B cos A q = angle between B and A MAGNETUIC FLUX: Magnetic flux is defined as the portion per unit area of the magnetic field penetrating perpendicularly through a surface. A q B The net magnetic flux through a closed surface is zero.

EXAMPLE: A uniform magnetic field, B, exists in space oriented horizontally to the right. At a certain point is a cube positioned so that its left face is perpendicular to the magnetic field. Each face of the cube has an area A. What is the magnetic flux through each face of the cube and the net magnetic flux through the cube? The flux through the left side is f1=-BA, A is outward and so is anti-parallel to B. The flux through the right side is f2=+BA, A is parallel to B. The flux through the other sides is 0, A is perpendicular to B. The net flux is the sum of the flux through each side, fnet = 0. The magnetic flux through any closed surface is always zero.

FARADAY’S LAW, INDUCED EMF: If a conducting loop or coil has through its encircled area a changing magnetic flux, then there will be an induced emf in the loop or coil. Df N= number of loops of wire = - E N D t In general: Df D D B A = + - q q w q cos A B cos sin BA D D D t t t In most problems the changing flux will depend on only one term.

mc mc BS t • EXAMPLE: A solenoid of length 10cm made with 1500 turns of wire carries a current of 12A that is steadily decreasing at the rate of 3A/s. Refer to the diagram for the initial direction of this current. Inside the solenoid is a coils whose center lies on the axis of the solenoid. The plane of the coil makes a 30O angle with the axis of the solenoid. The coil is made with 400 turns of wire, is circular with a radius of 3cm, and has a net resistance of 0.3 Ohms. • Calculate the following: •  the induced emf in the coil: the induced current in the coil: the magnetic dipole moment of the coil: the torque on the coil when Is is 2A:

I B I R v l FA x FM EXAMPLE: Sliding rod. Two parallel, horizontal, frictionless, conducting tracks are connected together at their left end by a wire with a resistor of resistance R. The separation between the rods isl. A conducting rod rests on the rod and is free to move along the rods. A uniform magnetic field, B, fills space and is directed vertically upward. The rod is pulled to the right with a constant velocity v. A conducting loop is form on the left with the rod on one side. Calculate the following: Tracks Rod Changing Area of Loop: Induced emf in loop Induced Current: Power Loss: Magnetic Force on Rod: Applied Force on Rod: Power Applied to Rod: Answers on next slide.

I B I R v l FA x FM EXAMPLE: Sliding rod. Two parallel, horizontal, frictionless, conducting tracks are connected together at their left end by a wire with a resistor of resistance R. The separation between the rods isl. A conducting rod rests on the rod and is free to move along the rods. A uniform magnetic field, B, fills space and is directed vertically upward. The rod is pulled to the right with a constant velocity v. A conducting loop is form on the left with the rod on one side. Calculate the following: Tracks Rod Changing Area of Loop: Induced emf in loop Induced Current: Power Loss: Magnetic Force on Rod: Applied Force on Rod: Power Applied to Rod:

= F F E B = eE evB E = e evB l = E Bvl MOTIONAL EMF: A conducting rod is moved through a magnetic field. Mobile electrons F experience a force that pulls them to one side of the rod. Thus one B end of the rod becomes electrically negative and the other becomes electrically positive. An electric field builds between the ends of the F rod. The electric field exerts a force on the electrons. Eventually E F F equals and no further charge moves. E B B l F B - - __ electron + F v E E rod See next slide for more details.

B E FB I - v FE MOTIONAL EMF: Uniform Magnetic Field - Mobile Electrons + Moving Rod Click to see each of the following: Effective current of electrons Magnetic force on electron Charge separation in rod Orientation of electric field Electric force on electrons

EXAMPLE: A metal bar of length 75cm is rolling down an incline at a steady speed of 3m/s. A uniform magnetic field of 0.5G exists in space perpendicular to the surface of the incline. What is the potential difference between the ends of the rod?

A q B Rotating Loop Df = - = E N NBA w sin q D t = = E E NBA w max o = E E sin q o THE AC GENERATOR: The induced emf causes an alternating current in the loop. The interaction between this current and the magnetic field is a torque that opposes the rotation of the loop. Thus an external source of energy will be required to keep the loop in rotation.

EXAMPLE: A coil is made to rotate inside a solenoid. The solenoid consists of 2500 turns of wire, has a length of 30cm, and carries a current of 6A. The coil is made with 500 turns of wire, is circular with a radius of 2cm, and has a net resistance of .75 Ohms. If the coil is rotating at 500 revolutions per minute, what is the maximum current in the coil?

E P Df Df = = E E N N = P P S S D D E E t t I I P P S S E E N E = = E E = P S S I I P S P S P N N N E P S P S Iron Core THE TRANSFORMER: Primary Coil Secondary A transformer consists of a primary coil (input voltage), a secondary coil (output voltage), and an iron core the coils are wrapped around. Coil N P N E S S Ideally: Magnetic field lines (in blue) from the primary coil are directed through the iron core to the secondary coil. The changing flux they cause in the secondary coil induces the emf in the secondary coil. This will not work if the current in the primary coil is constant.

EXAMPLE: The primary coil of a transformer has 2000 turns of wire. (a) If the maximum emf in the primary coil is 24V and 75V is required at the secondary coil, how many turns of wire are required in the secondary coil? (b) If the maximum current in the primary coil is 5A, what is the ideal maximum current from the secondary coil? A transformer whose secondary coil voltage is greater than that of the primary is called a step up transformer. If the secondary voltage is less than the primary voltage, it is called a step down transformer.

INDUCTANCE: V V Inductor E O E O When the switch is closed Back emf: Df D B D I E the voltage builds to . ö æ o = - = - = - m E N N A N n A The reason it does not ø è o D t D t D t E instantly become is o because the change in D I D I 2 N current produces a = - m = - E A L changing magnetic field, o D t D t l and an induced emf that E opposes . This emf is 2 N o back-emf called a and the = m L A self- process is known as o l induction . L is Inductance and has SI unit of Henry, H.

å å = = D D W W IL I å = = 2 D W L I I LI 1 2 = 2 W LI 1 2 STORED ENERGY: As current begins to flow through an inductor a magnetic field is created and increases in strength. Within the field energy is stored. In a short time D while the current is increasing an amount of energy is added D W t to the field. D = D = D W P t I t E D I æ ö D = D = t D W I L IL I è ø D t As the current increases more and more energy is stored until the current I stops increasing at . The total energy stored is : final

m 2 N A æ ö = = 2 2 W LI I o 1 1 è ø 2 2 l m NI æ ö = = W N AI NBAI o 1 1 è ø 2 2 l m NI l ö æ ö æ ( ) = = W NI BA BA o 1 1 è ø è ø 2 2 m l o 2 B ( ) = W Al 1 2 m o 2 B = u 1 2 m o STORED ENERGY: Energy density may also be calculated:

EXAMPLE: A circuit consists of a 20 Ohm resistor, 24V battery and an inductor in series. The inductor is a solenoid made of 100 turns of wire wrapped around a cylinder 2cm long with a radius of 5mm. Calculate the inductance of the solenoid and the energy density of the field inside.

MAGNETIC INDUCTION