FLUX THROUGH A SINGLE TURN COIL B The

FLUX THROUGH A SINGLE TURN COIL B The flux through the coil is given by Ф=AB Where A is the area of the coil But what happens to the flux if the coil turns (perpendicular to the field) ?

FLUX THROUGH A SINGLE TURN COIL B When the coil begins to turn. We can see that the flux through the coil becomes less and when it is horizontal it is zero. B

FLUX THROUGH A SINGLE TURN COIL θ Generalizing the result, taking a line normal to the plane of the coil (the dotted line) Where θ is the angle between the field and this normal line The flux through the coil is given by Ф=ABcosθ B

Flux Linkage When the coil has more than one turn the flux is equal to the sum of the fluxes through the individual turns. We call this the flux linkage though the whole coil.

FLUX LINKAGE B The flux linkage through the whole coil is given by NФ=NABcosθ Where N is the number of turns in the coil This formula is handy later when we consider the emf induced in a rotating coil.

Faraday’s Law • Accurate experiments show that the e. m. f. induced in a coil increases with the rate of change of magnetic flux through it. e. m. f. rate of change of flux The rate of change of flux is measured in Webers per second (Wm-1) ε

Faraday’s law of electromagnetic induction ε ε So combining these relationships ε The units of the SI system combine in such a way that the constant of proportionality is 1 ε The expression to the right of the = sign is just the rate of change of flux linkage

Faraday’s Law ε ε is the e. m. f. Induced (V) N is the number of turns on the coil ∆Φ is the change in flux though each turn of the coil. (Wb) ∆t is the time taken for the flux change. (s) Note that in this equation the total change in flux linkage in the coil is N∆Φ. Sometimes you may see this written as ∆NΦ. It follows that 1 weber is the flux linkage in a coil if an emf of 1 V falls evenly to zero in 1 second

Using Faraday’s Law A coil of 200 turns and 3 cm in diameter lies perpendicular to a uniform magnetic field with a flux density of 2 x 10 -2 T. The field falls evenly to 0 T in 1 s. Calculate the emf generated: . 1. Calculate the flux 2. through 1 turn of the coil 2 Now apply Faraday’s law ε V

Examples • A coil of 450 turns and 10 cm in diameter lies perpendicular between the poles of an electromagnet. The field rises evenly from 0 to 4 x 10 -3 T in 0. 5 s. Calculate the e. m. f generated: • A Helmholtz coil with 2000 turns of diameter 12 cm is set perpendicular to a uniform magnetic field of 0. 5 T. The field is reduced to 0 in 0. 2 seconds. Calculate the e. m. f generated.

The e. m. f. rotating coil • Now we can apply Faraday’s law to the rotating coil The Flux linkage is: NФ=NABcosθ

The e. m. f. rotating coil NФ=NABcosθ We need to find the rate of change of flux through the coil The angular speed of the coil is t θ

NФ=BANcosθ Becomes NФ=BANcosωt The e. m. f. is the rate of change of flux linkage And as δt → 0 ε ε Differentiation gives The minus sign is a consequence of Lenses Law and gives the direction of the emf. This can be ignored ε ε ε

The Rotating Coil The e. m. f. generated by a rotating coil is given by: ε B Note that this A SINUSOIDAL RELATIONSHIP. The e. m. f generated is varying:

The peak output of the generator is BANω when the sinωt = 1 ε ε T t This curve is characteristic of an a. c. generator

Questions • A circular coil with 300 turns and a diameter of 5 cm is rotated at a frequency of 5 Hz perpendicular to a magnetic field of 3. 0 x 10 -3 T. Calculate the e. m. f. generated at a quarter cycle. • A Helmholtz coil with 1000 turns of diameter 15 cm is set vertically in the Earth magnetic field and rapidly rotated with a frequency of 20 Hz. Calculate the peak e. m. f. you would expect to generate. (Take the horizontal component of the Earth’s magnetic field to be 3. 2 x 10 -5 T)