Faraday’s Law

Faraday’s law governs the phenomenon of electromagnetic induction. Electromagnetic induction refers to changing magnetic fields induced in any circuit by the Electromotive force or voltage in a conductor.

1.0What does Faraday's law state?

Faraday’s Law of Electromagnetic Induction - It states that the induced EMF (electromotive force) in any closed loop is always directly proportional to the rate of change of magnetic flux(B) through the loop. Mathematically, the same can be expressed as: 

$EMF=-N.\frac{d{\varphi }_B}{dt}$

Here: 

• ${\Phi }_B$=Magnetic flux

(${\varphi }_B$= BAcos$\theta$, here, B represents the magnetic field strength, A represents the area, and is the angle between the magnetic field and the normal to the surface area).

• $\frac{d{\varphi }_B}{dt}$= Rate of change of magnetic flux. 

According to Lenz's Law, the induced EMF is negative. It simply signifies that the induced current has its direction such that the changing magnetic flux is opposed.

Lenz's Law relates how the direction of the induced current, and therefore the EMF, is such that the inducing change in magnetic flux produces opposing force. 

2.0Faraday’s Laws of Electrolysis

Faraday's Laws of Electrolysis describe how the amount of substance that is deposited or dissolved at the electrodes is always proportional to the quantity of electricity used in electrolysis. Faraday’s Law of Electrolysis is divided into the first and second laws of electrolysis. 

1. First Law of Electrolysis: 

The mass of a substance that is deposited or dissolved in an electrode is directly proportional to the given quantity of electricity that passes through the electrolyte. Mathematically, it can be written as: 

$m=\frac{MQ}{F}$

Here:

• m = Mass of substance deposited/dissolved
• M = Molar mass of the substance
• Q = Total charge passed (in Coulombs)
• F = Faraday's constant (96,485 C/mol)

Example: When a current of 5 A is passed through a copper sulfate solution for 4 hours, calculate the mass of copper deposited at the cathode. The electrochemical equivalent of copper is 3.3×10⁻⁵ kg/C.

Solution: it is given that time t = 4h = 14400s and M = 3.3×10⁻⁵ kg/C

The total charge Q = It 

Q = 514400=72000C

Since the Electrochemical equivalent M is already given in the problem hence the formula to calculate mass m can be simplified as: 

$m=M\times Q$

$m=(3.3\times 10^{-5})72000=2.376Kg$

2. Second Law of Electrolysis: 

The masses of different substances deposited/dissolved by the same quantity of electricity are always directly proportional to their equivalent weights. In maths, it can be written as: 

$\frac{m_1}{m_2}=\frac{E_1}{E_2}$

Here:

• m₁ and m₂​ are the masses of two substances deposited,
• E₁ and E₂ are their respective equivalent weights.

Problem: If the electrochemical equivalent of silver is E_Ag=1.118×10⁻⁴ kg/C, and for copper E_Cu=3.3×10⁻⁵ kg/C, find the ratio of the equivalent masses of silver and copper.

Solution: Using the second law of electrolysis: 

$\frac{m_{Ag}}{m_{Cu}}=\frac{E_{Ag}}{E_{Cu}}$

$\frac{m_{Ag}}{m_{Cu}}$$=\frac{1.118\times 10^{-4}Kg/C}{3.3\times 10^{-5}kg/C}=3.39$

3.0Solved Examples 

Problem 1: A coil of 200 turns has a radius of 0.05 m and is placed in a magnetic field. The magnetic field strength changes from B₁=0.3 T to B₂=0.7 T in 0.2 s. The resistance of the coil is 5 Ω. Find the current induced in the coil.

Solution: Area of the coil $A=\pi r^2$

$A=3.14\times 0.05^2=7.854\times 10^{-3}{m}^2$

Change in magnetic flux = $\Delta {\Phi }_B$

${\Phi }_B={\Phi }_B(final)-{\Phi }_B(initial)$

${\Phi }_B=B_{2}A-B_{1}A$

$\Delta {\Phi }_B=5.498\times 10^{-3}-2.356\times 10^{-3}=3.142\times 10^{-3}Wb$

Induced EMF = $-\frac{d{\varphi }_B}{dt}=-\frac{3.142\times 10^{-3}}{0.2}$

Induced EMF = $–1.571\times 10^{-2}V=–15.71mV$

Induced current: using Ohm’s law 

$I=\frac{V}{R}=\frac{1.571\times 10^{-2}}{5}=3.14\times 10^{-3}A=3.14mA$

Problem 2: A coil of 500 turns and area A = 0.02 m² is placed in a uniform magnetic field. The magnetic field strength changes from B₁=0.2 T to B₂=0.6 T in a time interval of 0.5 s. Find the induced EMF in the coil.

Solution: Change in Magnetic flux: B

$\Delta {\Phi }_B={\Phi }_B(final)-{\Phi }_B(initial)$

$\Delta {\Phi }_B=B_{2}A-B_{1}A$

$\Delta {\Phi }_B=0.6\times 0.02-0.2\times 0.02$

$\Delta {\Phi }_B=0.012-0.004$

$\Delta {\Phi }_B=0.008=8\times 10^{-3}Wb$

$EMF=-\frac{d{\Phi }_B}{dt}$

$EMF=-\frac{8\times 10^{-3}}{0.5}$

$EMF=-1.6\times 10^{-2}V$

Problem 3: A coil with N=500 turns, and area A=0.05 m² is exposed to a magnetic field. If the magnetic field strength changes from B₁=0.1 T, and the induced EMF is found to be 0.5 V, calculate the required magnetic field strength.

Solution: Change in magnetic flux =

$\Delta {\Phi }_B=B_{2}A-B_{1}A$

$\Delta {\Phi }_B=(B_2-B_1)A$

$EMF=-N\frac{d{\varphi }_B}{dt}$

$0.5=-500\times \frac{(B_2-0.1)0.05}{0.2}$

$1=-500(B_2-0.1)0.05$

$B_2=0.4T$