Faraday's Law

Lenz's Law

Direction of current flow is so as to oppose change in magnetic flux.

$\Phi_B >o$

$\frac {d\Phi_B}{dt}>0$

$\varepsilon = -\frac{d\Phi_B}{dt}<0$

Uniform magnetic field $\vec B$

$\vec B$ is increasing with time, $\frac {dB}{dt}$ = 0.04 T/s

Radius of the conducting loop r = 5cm

Let the resistances of the loop R = 5$\Omega$

$\varepsilon =-\frac{d {\Phi}_B}{d t}$

${\Phi}_B = B A = B \pi r^2$

$\varepsilon =-\pi r^2 \frac{d B}{d t}$

$=-\pi \times 25 \times 10^{-4} \times 0.04$

= 0.314 mV

Induced current $I=\frac{\varepsilon}{R}=\frac{0.314 \times 10^{-3}}{5} \simeq 63 \mu A$

A very long solenoid $S_1$ with $N_1$ turns/unit length.

Current through $S_1$ = $I_1$

Magnetic fields within $S_1$ = $\mu_0 N_1 I_1$

Uniform within the solenoid

Inner small solenoid $\S_2$

Total number of turns $=N_{2 T}$

Radius of $S_2 = R_2$

Magnetic flux per turn = B.A

passing through $S_2 = \mu_0 N_1 I_1 \cdot \pi R_2^2$

Total flux through $S_2 = \mu_0 N_1 I_1 N_{2 T} \pi R_2^2$

$\varepsilon =-\frac{d \phi_ B}{d t} =-\mu_ 0 N_ 1 N_ {2 T} \pi R_ 2^2 \frac{d I_1}{d t}$

$N_ 1$ = 100 turns/cm, $I_1$ = 1 A

$N_ {2}$ = 100, $R_2$ = 1 cm

Current $I_1$ changes from 1A to zero in 10ms

$\frac{d I_1}{d t} = -\frac{1}{10^{-2}}$ = - 100 A/s

Induced EMF

$\varepsilon = + 4 \pi \times 10^{-7} \times 10^{4} \times 100 \times \pi \times 10^{-4} \times 100 \simeq + 39.5 mv$

N : Number of turns per unit length

Area of solenoid = $\pi r^2$

very long solenoid B = $\mu_0 N I$

Flux through the conducting loop = $B \pi r^2$ = $\mu_0 N I \pi r^2$

$\varepsilon = -\frac{d \Phi_B}{d t}$

$=-\mu_0 N A \frac{d I}{d t}$

$\varepsilon = \oint \vec{E} \cdot \vec{dl} =-\frac{d \Phi_B}{d t}$

$= - \mu_0 N A \frac{d I}{d t}$

For $\frac{d I}{d t}>0$

$\varepsilon<0$

$\varepsilon=\oint_C \vec{E} \cdot \vec{dl}=-\frac{d}{d t} \int_A \vec{B} \cdot d \vec{A}$

$\oint \vec{\varepsilon} \cdot \overrightarrow{d l} = 2 \pi R \varepsilon=-\mu_0 N A \frac{d I}{d t}$

$\varepsilon=-\frac{\mu_0 N A}{2 \pi R} \frac{d I}{d t}$

$\oint \vec{E} \cdot \vec{dl} \neq 0$

Electrostatic field $\oint \vec{E} \cdot \vec{dl}=0$

Number of turns $N=1000 / m$ per unit length

$\frac{d I}{d t}=100 A / s$

Area of the solenoid $=\pi \times 25 \times 10^{-4} {m}^2$ (radius of solenoid $=5 \mathrm{~cm}$)

Induced electric field at R = 10cm ?

$\varepsilon =-\frac{\mu_0 N A}{2 \pi R} \frac{d I}{dt}$

$=\frac{-4 \pi \times 10^{-7} \times 1000 \times \pi \times 25 \times 10^{-4} \times 100}{2 \pi \times 0.1}$

$\simeq-1.57 \times 10^{-3} \mathrm{~V} / m z4$

For $\frac{d I}{d t}>0$, $\varepsilon< 0$

$\varepsilon=\oint_ C \vec{\varepsilon} \cdot \vec{dl}$ =-$\frac{d}{d t} \int_A \vec{B} \cdot d \vec{A}$

$\oint \vec{\varepsilon} \cdot \overrightarrow{d l}=2 \pi R \varepsilon=-\mu_0 N A \frac{d I}{d t}$

$\varepsilon=-\frac{\mu_0 N A}{2 \pi R} \frac{d \underline{I}}{d t}$

$\vec{F} =q \vec{v} \times \vec{B}$

=q v B

q E = q v B

E =v B

Potential differnce between the end = v B L

Work done per unit time = Force $\times$ velocity

$=\frac{B^2 L^2 v}{R} \times v$

$=\frac{B^2 L^2 v^2}{R}$

Joule heating = $I^2 R$

=$(\frac{v B L}{R})^2 R$

$=\frac{B^2 L^2 v^2}{R}$