Electricity and Magnetism

Electricity and Magnetism are two key areas of physics that study electric charges, electric and magnetic fields, and how they interact. Electricity focuses on how electric charges behave, how current flows, the role of voltage, and how circuits work. Magnetism, on the other hand, looks at magnetic fields and how they affect moving charges. Together, these two topics form the core of electromagnetism, which helps us understand everything from how electrical circuits work to the nature of electromagnetic waves.

1.0Coulomb’s Law

The electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance separating them. This force always acts along the line joining the two charges.

$F=K\frac{q_1{q}_2}{r^2}$

$k=\frac{1}{4\pi {ϵ}_0}=9\times 10^{9}N{m}^2{C}^{-2}$=Electrostatic constant or Coulomb’s Constant

2.0Electric Field, Electric Potential and Electric Potential Energy

Electric Field: An electric field is the region around a charge or charge distribution where another charge experiences an electric force.

$\vec{E}=\frac{\vec{F}}{q_0}$ SI unit : N/C or V/m

Electric potential: It is defined as the work done by an external force in moving a unit positive charge from a reference point to a specific location without changing its kinetic energy.

$V_p=\frac{(W_{\infty \to p}{)}_{ext}}{q}$         (K=0)

Electric Potential Energy: It is the work done to move a charge from infinity to its current position without altering its kinetic energy.

$U=\frac{kQq}{r}$

3.0Electric Field Lines and Flux

Electric Field Lines: Electric field lines are imaginary lines, straight or curved, that represent the direction and strength of an electric field. The tangent at any point on a field line shows the direction of the field at that point.

Electric Flux: This physical quantity is used to measure strength of electric field and it is defined as the total number of electric field lines passing through an area.

$\varphi =\vec{E}.\vec{A}$

$\varphi =EACos\theta$

4.0Gauss’s Law

According to this law the total electric flux () through any closed surface (S) in free space is equal to $\frac{1}{{ϵ}_0}$ times the total electric charge (q) enclosed by the surface.

$\varphi =\oint \vec{E}.\overrightarrow{dS}=\frac{Q_{enclosed}}{{ϵ}_0}$

Applications of Gauss Law

(1).Electric field Intensity due to infinitely long wire, $E=\frac{\lambda }{2\pi {ϵ}_{0}r}=\frac{2K\lambda }{r}$

(2).Electric Field due to Uniformly Charged Infinite Sheet

(A).Non Conducting Sheet $E=\frac{\sigma }{2{ϵ}_0}$   (B) Conducting sheet or Metal Plate $E=\frac{\sigma }{{ϵ}_0}$

(3).Electric field due to uniformly charged long cylindrical pipe/cylindrical shell

Case 1.Electric field at any point outside the cylinder(r>R)    

$E=\frac{\sigma R}{{\epsilon }_{0}r}$

Case 2.For the point lying on the surface(r≈R)

$E=\frac{\sigma }{{\epsilon }_0}$

Case 3.For the point inside the surface(r<R)

$E_{\text{inside}}=0$

(4). Electric Field due to the charged conducting sphere or charged thin shell

(a). Electric Field at any point outside the sphere (r>R)

$E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{kq}{r^2}$

(b). For any point lying on the surface of sphere (r=R)

$E_s=\frac{kq}{R^2}=\frac{q}{4\pi {ϵ}_0{R}^2}=\frac{\sigma }{{ϵ}_0}$ $(\therefore \sigma =\frac{q}{4\pi R^2}$

(c). Electric Field at any point Inside the sphere(r<R)

In this case charge enclosed by the gaussian surface is zero

$E_{inside}=0$

(5).Electric Field due to uniformly charged non conducting sphere(solid sphere)

(a). Electric field at any point outside the sphere (r>R): $E=\frac{kq}{r^2}$

(b). Electric field at any point lying on the surface of sphere(r=R) $E_S=\frac{\rho R}{3{ϵ}_0}$    (∴ $\rho$ is volume charge density)

( c). Electric field at any point inside the sphere(r<R) $E_{inside}=\frac{1q}{4\pi {ϵ}_0{R}^3}r=\frac{\rho }{3{ϵ}_0}(r)$

5.0Electrical Capacitance(C) and Capacitors

Electrical Capacitance(C)

• It shows the capacity of a conductor to store electric charge. $C=\frac{Q}{V}$
• Capacitance of Parallel Plate Capacitor  $C=\frac{{ϵ}_{0}A}{d}$
• Capacitance Of Spherical Capacitor$C=4\pi {ϵ}_{0}r$
• Capacitance of Spherical Capacitor $C=4\pi {ϵ}_0(\frac{ab}{b-a})$
• Capacitance of cylindrical Capacitor $C=\frac{2\pi {ϵ}_{0}L}{lo{g}_e(\frac{b}{a})}$

Capacitors

Definition: It is an electrical component to store electric energy in the form of charge.

Capacitors in series : $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

Capacitors in parallel: $C=C_1+C_2+C_3+.......C_n$

6.0Electric Current and Ohm’s Law

Electric Current: Electric current (I) is referred to as the rate of flow of any charge Q through a conductor. The device used for producing electric current (I) is called an Electric Generator. The electric current formula is written as: $I=\frac{Q}{t}$

Ohm’s Law: In Physical quantities like temperature, pressure, volume, length, cross-section or nature of the material kept constant then current through a conductor is directly proportional to potential difference applied across it. This is called Ohm’s Law.

$V\propto I$

$\frac{V}{I}=\text{Constant}=\text{R(Resistance)}$

$V\propto I$

$V=IR$

R is called the resistance of the conductor.

7.0Combination of Resistors

Resistors in Series

• The total resistance is the sum of individual resistances: R_eq = R₁ + R₂ + …..
• Current: Each resistor has the same current flowing through it.
• Voltage: The total voltage equals the sum of the voltage drops across all the resistors.

Resistors in Parallel

• The total resistance is given by:$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+........$
• Current: The overall current is obtained by adding the currents through each resistor
• Voltage: The same voltage is applied to all resistors.

8.0Cells and its Combination

A cell is a device that provides the necessary potential difference to maintain a continuous flow of current in an electric circuit. It consists of two electrodes, typically rods or plates, which are immersed in a chemical solution known as the electrolyte. 

Combination of Cells in Series

$E_{eq}.=E_1+E_2+E_3+.......E_n$

$r_{eq}=r_1+r_2+r_3+.............r_n$

$R_{eq}=r_1+r_2+r_3+...........r_n+R$

Current $i=\frac{E_{eq}}{r_{eq}+R}$

If all n cells are identical then $i=\frac{nE}{nr+R}$

Combination of Cells in Parallel

$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}+...}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}....}$

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+......$

9.0Kirchhoff’s Laws 

Kirchhoff’s First Law Statement

• This law is also known as junction law or current law (KCL).According to this-In an electric circuit, the net sum of the currents gathering at any junction in the circuit is zero.
• Sum of the currents arriving the junction is equal to sum of the currents departing the junction

$\sum i=0$

$i_2+i_3+i_4-i_1-i_5=0$

$\sum i=0$

Kirchhoff’s Second Law Statement

• It is also known as loop rule or voltage law (KVL) and according to it in any closed circuit the algebraic sum of e.m.f. and algebraic sum of potential drops is zero. 

$\sum IR+\sum E=0$

10.0Heating Effect of Current

• It states that the heat (H) produced by a current-carrying conductor is equal to the product of the resistance(R), the square of the current (I), & the time (t) for which the current flows.  $H=I^{2}Rt$
• The SI unit for the heating effect of electric current is joule. 

11.0Magnetic Effects of Current

Biot-Savart Law

The magnetic field at a point is directly proportional to the current, element length, and sine of the angle, and inversely proportional to the square of the distance.                

$dB=\frac{{\mu }_0}{4\pi }\frac{IdlSin\theta }{r^2}$

${\mu }_0=4\pi \times 10^{-7}Tm{a}^{-1}$

Application of Biot-Savart Law

1. Magnetic Field due to Thin Wire of Finite Length

$B=\frac{{\mu }_{0}I}{4\pi a}[sin{\varphi }_1+Sin{\varphi }_2]\text{inwards}$

2. Magnetic Field due to Infinite Straight Wire

$B=\frac{{\mu }_{0}I}{2\pi a}$

3. Magnetic Field due to Semi Infinite Straight Wire

Case-1

${\varphi }_1\to 90^{o}or{\varphi }_2\to 0°$

$B={\frac{{\mu }_{0}I}{4\pi a}}^{-\otimes }$

Case-2.

${\varphi }_1\to 90^{o}or{\varphi }_2\to \theta$

$B=\frac{{\mu }_{0}I}{4\pi a}(1+sin\theta {)}^{-\otimes }$

Case-3.

${\varphi }_1\to 90^{o}or{\varphi }_2\to -\theta$

$B=\frac{{\mu }_{0}I}{4\pi a}(1-sin\theta {)}^{-\otimes }$

Case-4.

${\varphi }_1\to -\theta or{\varphi }_2\to 90^0$

$B=\frac{{\mu }_{0}I}{4\pi a}(1-sin\theta {)}^{-\otimes }$

12.0Ampere’s Circuital Law

Line integral of magnetic field along any closed loop is equal to o times the net current crossing the surface bounded by the loop.

$\oint \vec{B}.\overrightarrow{dl}={\mu }_0{I}_{enc}$

13.0Force in Magnetic Fields

Magnetic Force on Moving Charge

$\vec{F}=\vec{q}(\vec{v}\times \vec{B})=qvBsin\theta$

Magnetic force depends on angle between $\vec{v}$ and  $\vec{B}$

Magnetic Force on a Current Carrying Wire

$\overrightarrow{dF}=I(\overrightarrow{dl}\times \vec{B})$

$\vec{F}=I(\overrightarrow{L_{eff}}\times \vec{B})L_{eff}$ is the displacement vector from starting point of current to end

point of current.

14.0Magnetic Moment 

Magnetic Moment of Current Carrying Coil (Loop)

Magnetic Moment $\vec{M}=NI\vec{A}$

15.0Instruments: Galvanometer, Voltmeter, Ammeter

1. Moving Coil Galvanometer

The galvanometer has a coil with many turns, free to rotate in a uniform radial magnetic field. A soft iron core strengthens and radializes the field, while a spiral spring resists the coil's rotation.

• A current bearing coil in a magnetic field experiences a torque.  $\tau =NIABsin\theta$
• The spring S provides a counter torque C that balances the magnetic torque, ${\tau }^{\prime }=C\varphi$
• In equilibrium,$C\varphi =NIAB\Rightarrow I=\frac{C}{NAB}\varphi$, $I\propto \varphi$

It means the deflection produced is proportional to the current flowing through the galvanometer.

2. Voltmeter

• A voltmeter is joined in parallel to the current carrying wire, to measure the potential difference between two points.
• To convert a galvanometer into a voltmeter, a high resistance is connected in series with it.
• The resistance of the voltmeter is very high and it is infinite for an ideal voltmeter.
•  So ideal voltmeter open circuit

$H=(\frac{V-V_g}{V_g})G=(\frac{V}{V_g}-1)G$

$H=(n-1)G$             where $n=\frac{V}{V_g}$

V=Range of Voltmeter

Vg=Range of Galvanometer

3. Ammeter

• An ammeter is connected in series with a current carrying wire to measure current passing through it.
• To convert a galvanometer into an ammeter a very small resistance is connected in parallel to the galvanometer called SHUNT.
• Resistance of an ammeter is very small and it is zero for ideal ammeter i.e. ideal ammeter behaves like conducting wire.
• Value of shunt.

$\Rightarrow S=[\frac{i_g}{i-i_g}]G$

$\Rightarrow S=[\frac{1}{\frac{i}{i_g}-1}]G\Rightarrow S=[\frac{G}{n-1}]$$wheren=\frac{i}{i_g}$

G– resistance of galvanometer

ig-range of galvanometer or current required to produce full deflection

i-Range of ammeter [Max. current can be measured.]

16.0Electromagnetic Induction

Faraday’s First Law-An electromotive force (EMF) is induced in a conductor whenever it is exposed to a changing magnetic field.

Faraday’s Second Law-The induced emf in a coil is equal to the rate of change of flux linkage.

$|e|=\frac{d\varphi }{dt}$, Induced EMF ∝ Relative Velocity

Lenz Law-The induced EMF's polarity is such that it generates a current opposing the change in magnetic flux responsible for its creation.

$e=-\frac{d\varphi }{dt}$ ( Negative sign denotes opposition)

17.0Self Induction(L)

When current through the coil changes, with respect to time then magnetic flux linked with the coil also changes with respect to time. Due to this an emf and a current is induced in the coil. According to Lenz law, induced current opposes the change in magnetic flux. This phenomenon is called self-induction and a factor by virtue of which the coil shows opposition to change in magnetic flux called self-inductance of the coil. 

Case 1. Current through  the coil is constant

$L=\frac{N\varphi }{I}=fracNBAI=\frac{{\varphi }_{Total}}{I}$

Case 2. Induced EMF in Self Induction

$e_s=\frac{-LdI}{dt}$

18.0Mutual Induction(M)

Whenever current is passing through the primary coil or circuit, changes with respect to time then magnetic flux in neighboring secondary coil or circuit will also change with respect to time. According to Lenz Law for opposition of flux change an emf and a current induced in the neighboring coil or circuit. This phenomenon is called 'Mutual induction'.

$M=\frac{N_2{\varphi }_2}{I_1}=\frac{N_2{B}_1{A}_2}{I_1}=\frac{({\varphi }_T{)}_S}{I_P}$

19.0Alternating Current

Definition: An alternating quantity (such as current or voltage) is one whose magnitude continuously varies with time between zero and a maximum value, while its direction periodically reverses.

$I=I_{0}Sin\omega t$

$I=I_{0}Cos\omega t$

• AC Circuit Containing Resistor

$I=\frac{E_{0}Sin\omega t}{R}=I_{0}Sin\omega t$

$I_0=\frac{E_0}{R}$=Peak or Maximum Value of Current

• AC Circuit Containing Inductor=$I=I_{0}Sin(\omega t-\frac{\pi }{2})$
• AC Circuit Containing Capacitor=$I=I_{0}Sin(\omega t+\frac{\pi }{2})$
• AC Circuit Containing Series LCR Circuit

$I=\frac{E}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}$

$Tan\varphi =\frac{X_L-X_C}{R}$

• Power Associated with series LCR Circuit, $P_{avg}=V_{rms}{I}_{rms}Cos\varphi$

20.0Charging and Discharging Formulas for LR and RC Circuits

21.0Sample Questions on Electricity And Magnetism

Q-1. A particle of mass m   and charge $q_1$​ is moving in a circular orbit of radius r under the electrostatic attraction of a fixed point charge $q_2$ (located at the center of the circle).Determine the speed of the particle in its orbit and calculate the time period of its revolution.

Solution:

$\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}=mr{\omega }^2=\frac{4{\pi }^{2}mr}{T^2}$

$T^2=\frac{(4\pi {ϵ}_o)r^2(4{\pi }^{2}mr)}{q_1{q}_2}$$orT=4\pi r\sqrt{\frac{\pi {ϵ}_{0}mr}{q_1{q}_2}}$

$\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}=\frac{m{v}^2}{r}\Rightarrow v=\sqrt{\frac{q_1{q}_2}{4\pi {ϵ}_{0}mr}}$

Q-2. The current through a wire relies on time as $i=i_o+\alpha sin\pi t$ where $i_o=10A$ and $\alpha =\frac{\pi }{2}A$, Find the charge crossed through a section of the wire in 3 seconds, and average current for that interval.

Solution:

$I=\frac{dq}{dt}\Rightarrow dq=Idt$

$q={\int }_0^3(i_0+\alpha sin\pi t)dt$

$q=[i_{0}t+\frac{\alpha }{\pi }(-cos\pi t){]}_0^3$

$q=3i_o+\frac{2\alpha }{\pi }=31C$

Average current is :

$I_{avg}=\frac{q}{Deltat}=\frac{31C}{3s}\Rightarrow I_{avg}=\frac{31}{3}A$

Q-3. An infinite number of capacitors of capacitance C, 4C, 16C... are connected in series then what will be their resultant capacitance?

Solution:

Let the equivalent capacitance of the combination= $C_{eq}$

$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{4C}+\frac{1}{16C}+..........\infty =[1+\frac{1}{4}+\frac{1}{16}+..........\infty ]\frac{1}{C}$

this is G.P series

$\Rightarrow S_{\infty }=\frac{a}{1-r}$ first term , a=1, common ratio=$\frac{1}{4}$

$\frac{1}{C_{eq}}=\frac{1}{1-\frac{1}{4}}\times \frac{1}{C}\Rightarrow C_{eq}=\frac{3}{4}C$

Q-4. A charged body of mass m  and charge  q  is initially at rest at the origin. It enters a region where a constant electric field $\vec{E}=E_o\hat{i}$ and a constant magnetic field $\vec{B}=B_o\hat{i}$ are both directed along the x-axis. At time t=0, the body is given an initial velocity $\vec{v}=v_o\hat{j}$ (along the y-axis).After how much time will the magnitude of its velocity become $2v_0$​?

Solution:

A charged particle will move in a Helical path.

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

$\vec{a}=\frac{qE}{m}\hat{i}+\frac{q{v}_o\times B}{m}$

$v_x=\frac{qE}{b}.t$

$v=\sqrt{v_x^2+v_o^2}$

$2v_o=\sqrt{(\frac{qE}{m}t{)}^2+v_o^2}$

$\sqrt{3}{v}_o=\frac{qE}{m}t$

$t=\frac{\sqrt{3}m{v}_o}{qE}$

Q-5. If power factor of a R-L series circuit is $\frac{1}{2}$ when applied voltage $v=100sin100\pi t$ volt and resistance of circuit is $200\Omega$ then calculate the inductance of the circuit.

Solution:

$cos\varphi =\frac{R}{Z}\Rightarrow \frac{1}{2}=\frac{R}{Z}\Rightarrow Z=2R\Rightarrow \sqrt{R^2+X_L^2}=2R\Rightarrow X_L=\sqrt{3}R$

$\omega L=\sqrt{3}R\Rightarrow L=\frac{\sqrt{3}R}{\omega }=\frac{\sqrt{3}\times 200}{100\pi }=\frac{2\sqrt{3}}{\pi }H$