====== IX. Electromagnetic induction ====== ===== - Faraday's law ===== {{:courses:phy102:9-3.jpeg?nolink|}} Magnetic flux is defined as $$ \Phi_m = \int_S \vec{B} \cdot d\vec{A} = \int_S \vec{B} \cdot \hat{n} dA $$ where $\hat{n}$ is the normal vector to the area $\vec{A}$. Unit is weber, Wb. 1 Wb = 1 T m$^2$. Michael Faraday observed the following phenomena. {{:courses:phy102:9-1.jpeg?nolink|}} {{:courses:phy102:9-2.jpeg?nolink|}} Faraday's law says that the induced emf (electromotive force, voltage) in a circuit $$ \varepsilon = -\frac{d\Phi_m}{dt} $$ that is the emf is the negative of the rate of change of the magnetic flux through the circuit. {{:courses:phy102:9-4.jpeg?nolink|}} If there are multiple turns like above then $$ \varepsilon = -N \frac{d\Phi_m}{dt}. $$ ===== - Lenz's law ===== {{:courses:phy102:9-5.jpeg?nolink|}} //The direction of the induced emf drives current around a wire loop to always oppose the change in magnetic flux that causes the emf.// ===== - Motional emf ===== {{:courses:phy102:9-6.jpeg?nolink|}} Magnetic flux varies with field, area and the angle between field and area. Variation of any one of these will induce an emf. {{:courses:phy102:9-7.jpeg?nolink|}} The area of MNOP $A=lx$ and the flux $\Phi=BA=Blx$. The induced emf $$ \varepsilon = -\frac{d\Phi}{dt} = Bl\frac{dx}{dt} = Blv $$ and the current induced in the circuit $$ I = \frac{\varepsilon}{R} = \frac{Blv}{R}. $$ From the energy point of view, $\vec{F}_a$ produces power $P=F_a v$: $$ F_a v = IlBv = \frac{Blv}{R} Blv = \frac{l^2 B^2 v^2}{R}. $$ The power dissipated $$ P = I^2 R = \frac{l^2 B^2 v^2}{R}. $$ The produced and dissipated power are equal. {{:courses:phy102:9-9.jpeg?nolink|}} ===== - Induced electric field ===== Faraday's law can rewritten in terms of the induced electric field. $$ \varepsilon = \oint \vec{E} \cdot d\vec{l} = - \frac{d\Phi_m}{dt}. $$ Note that for the electrostatic case $\oint \vec{E} \cdot d\vec{l}=0$. ===== - Electric generators and back emf ===== {{:courses:phy102:9-10.jpeg?nolink|}} Calculate the emf induced when the call is rotated from $\theta=0^\circ$ to $90^\circ$: $$ \varepsilon = -\frac{d\Phi}{dt} = NBA \sin\theta \frac{d\theta}{dt}. $$ Here $A=\pi r^2$, $N=200$, $r=5$ cm, $B=0.8$ T, $d\theta=\pi/2$ and $dt=15$ ms. {{:courses:phy102:9-11.jpeg?nolink|}} In the above diagram, emf is produced on both side of the loop, total emf $$ \varepsilon = 2Blv\sin\theta = 2Blv\sin(\omega t) $$ where $\omega$ is the angular velocity. So $v=r\omega$ and $r=w/2$ if $w$ is the width of the loop. So $$ \varepsilon = 2Bl \frac{w}{2}\omega \sin(\omega t) = (lw)B\omega \sin(\omega t) = AB\omega \sin(\omega t). $$ If there are $N$ loops just multiply by $N$. We can write $$ \varepsilon = \varepsilon_0 \sin(\omega t) $$ where $\varepsilon_0 = NBA\omega$ as you see below. {{:courses:phy102:9-12.jpeg?nolink|}} The frequency of the oscillation of this alternating current is $f=\omega/(2\pi)$ and the period $T=2\pi/omega$. {{:courses:phy102:9-14.jpeg?nolink|}}