Magnetic field B = 1 T
Mass of the proton, m_p = 1.67 x 10⁻²⁷kg
Charge of the proton, q = 1.60 x 10⁻¹⁹C
\(f=\frac { qB }{ { { 2\pi m }_{ p } } } =\frac { \left( 1.60\times { 10 }^{ -19 } \right) \left( 1 \right) }{ 2\left( 3.14 \right) \left( 1.67\times { 10 }^{ -27 } \right) } \)
= 15.3 x 10⁶ Hz = 15.3 MHz

Given that I = 1 A and radius r = 1 m
B_straightwire =  \(=\frac { { \mu }_{ ° }I }{ 2\pi r } =\frac { 4\pi \times { 10 }^{ -7 }\times 1 }{ 2\pi \times 1 } =2\times { 10 }^{ -7 }T\)
But the Earth’s magnetic field is B_earth \(\sim { 10 }^{ -5 }T\)
So, B_straightwire is one hundred times smaller than B_Earth.

The amplitude of the electric field, E₀ = 3 x 10⁴ NC^-1
The amplitude of the magnetic field, B₀ = 2 x 10^-4 T. Therefore, speed of the electromagnetic wave in a medium is
v = \(\frac { 3\times { 10 }^{ 4 } }{ 2\times { 10 }^{ -4 } } \)  = 1.5 x 10⁸ ms^-1.

Dielectric constant (relative permittivity of the medium) is ε_r = 2.25
Magnetic permeability is μ_r = 2.5
Refractive index of the medium,
n = \(\sqrt { { \varepsilon }_{ r }{ \mu }_{ r } } =\sqrt { 2.25\times 2.5 } \) = 2.37

Case (i):
di₁ = 10 – 2 = 8 A; dt = 0.4 s;
ε₂ = 60 x 10^-3V
Case(ii):
di₁ = 16 – 4 = 12 A; dt = 0.03 s
(i) Mutual inductance between the coils.
\({ M }=\frac { { \epsilon }_{ 2 } }{ \frac { { di }_{ 1 } }{ dt } } \)
\(=\frac { 60\times { 10 }^{ -3 }\times 0.4 }{ 8 } \)
\({ M }=3\times { 10 }^{ -3 }H\)
(ii) Induced emf in the second coil due to the rate of change of current in the first coil is
\({ \epsilon }_{ 2 }={ M }=\frac { { di }_{ 1 } }{ dt } \)
\(=\frac { 3\times { 10 }^{ -3 }\times 12 }{ 0.03 } \)
ε₂ = 1.2V

N = 40 turns; B = 2 Wb m^-2
A = 200 cm² = 200 x 10^-4 m²;
Initial flux, \(\Phi_i\) = BA cos\(\theta\)
= 2 x 200 x 10^-4 x cos60^o
since θ = 90°− 30°= 60°
\(\Phi_i\)= 2 x 10^-2 Wb
Final flux, \(\Phi_f\) = BA cos\(\theta\)
= 2 x 200 x 10^-4 x cos0^o since \(\theta\) = 0^o
\(\Phi_f\) = 4 x 10^-2Wb
Magnitude of the induced emf is
\(ε =N\frac { d{ \Phi }_{ B } }{ dt } \)
\(=\frac { 40\times (4\times { 10 }^{ -2 }-2\times { 10 }^{ -2 }) }{ 0.2 } =4V\)

The electric field on the surface of the sphere(by Gauss law) is given by
E = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { R }^{ 2 } } \)
The potential on the surface of the hollow metallic sphere is given by
V = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { R } } \) = ER
Since V_max = E_maxR
Here E_max = 3 x 10⁶ Vm^-1. So the maximum potential difference created is given by
V_max = 3 x 10⁶ x 0.5
= 1.5 x 10⁶V (or) 1.5 million volt.

R = 10 Ω, I = 5 A, t = 5 minutes = 5 x 60 s
H = I² R t
= 5² x 10 x 5 x 60
= 25 x 10 x 300
= 25 x 3000
= 75000 J (or) 75 kJ

R₀ = 3 Ω, T = 100^oC, T₀ = 20^oC
α = 0.004/^oC, R_T = ?
R_T = R₀(1 + α(T - T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
R₁₀₀ = 3.96 Ω

E = 570 N C^-1, e = 1.6 x 10^-19 C,
m = 9.11 x 10^-31 kg and a = ?
F = ma = eE
\(a=\frac { eE }{ m } =\frac { 570\times 1.6 \times { 10 }^{ -19 } }{ 9.11\times { 10 }^{ -31 } } \)
\(=\frac { 912\times { 10 }^{- 19 }\times { 10 }^{ 31 } }{ 9.11 } \)
= 1.001 x 10¹⁴ ms^-2

B = ∆m x c²
∆m = \(\frac{B}{c^2}\)
N M_n + Z M_p - M(N,Z) = \(\frac{B}{c^2}\)
N (N,Z) = N M_n + ZM_p - \(\frac{B}{c^2}\)

\(L=\frac{nh}{2\pi}=\frac{4h}{2\pi}=\frac{2h}{\pi}\)

I = l₁ + l₂ + 2\(\sqrt{I_1I_2}\)cos θ
If cos θ = cos 0 = l, I is max
= I+ 4I + 2\(\sqrt{41^2}\) cos 0
= 5I + 4I = 91
If cos π = -1, I is min
I_min = I + 4I + 2\(\sqrt{41^2}\) cos π
= 5I + 4I(-1)
= 5I + 4I = I
(I_max, I_min)= (9I, I)

Apparent depth = 3 + 5 = 8 cm
Real depth = thickness of the slab = t
n = 1.5
\(n=\frac{Real \ depth}{Apparent \ depth}\)
\(\therefore 1.5=\frac{t}{8}\)
t = 1.5 x 8
t = 12 cm

Dipole moment, \(\vec{p}_m=n\times I\times\vec{A}\)
\(\vec{p}_m\) = 50 x 3 x 3.14 x 25 x 10^-4 ≈ 1.2 A m²

\(\phi_B =10 t^2-50 t+250  \)
\(e  =\frac{-d \phi_B}{d t}  \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50) 
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V

Total power = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A

Resistors in parallel:
(i) Resistors are in parallel when they are connected across the same potential difference as shown in figure (a).
(ii) In this case, the total current I that leave the battery split into three separate components.
Let I₁, I₂ and I₃ be the current through the resistors R₁, R₂, and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I₁ + I₂ + I₃  .....(1)
(iii) Since the voltage across each resistor is the same, applying Ohm's law to each resistor, we have
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } ,{ I }_{ 2 }=\cfrac { V }{ { R }_{ 2 } } ,{ I }_{ 3 }=\cfrac { V }{ { R }_{ 3 } } \) 
Substituting these values in equation (1), we get,
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } +\cfrac { V }{ { R }_{ 2 } } +\cfrac { V }{ { R }_{ 3 } } =V\left[ \cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R }_{ 2 } } +\cfrac { 1 }{ { R }_{ 3 } } \right] \)
\(I=\cfrac { V }{ { R }_{ p } } \)
\(\cfrac { 1 }{ R_{ P } } =\cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R_{ 2 } } } +\cfrac { 1 }{ R_{ 3 } } \)  

Here R_P is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination.
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Biot-Savart's law states that the magnitude of magnetic field \(d\vec { B } \) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length of element \(\vec { dl } \)
(iii) directly as the sine of the angle (say,θ) between \(\vec { dl } \) and \(\hat { r } \).
(iv) inversely as the square of the distance between the point P and length element \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)

Consider a conducting rod of length l moving with a velocity v towards left on a rectangular fixed metallic framework as shown in Figure. The whole arrangement is placed in a uniform magnetic field \(\vec { B } \) whose magnetic lines are perpendicularly directed into the plane of the paper.
As the rod moves from AB to DC in a time dt , the area enclosed by the loop and hence the magnetic flux through the loop decreases.

The change in magnetic flux in time dt is
\(d{ \Phi }_{ B }=B\) x change in area (dA)
= B x Area ABCD
\(d{ \Phi }_{ B }\)= Blvdt (Since area ABCD = I (vdt))
(or) \(\frac { d{ \Phi }_{ B } }{ dt } =Blv\)
As a result of change in flux, an emf is generated in the loop. The magnitude of the induced emf is
ε = \(\frac { d{ \Phi }_{ B } }{ dt } \)
ε = \(Blv\)
This emf is called motional emf.

Consider a long solenoid of length Iand cross-sectional area A. Let n be the number of turns per unit length (or turn density) of the solenoid. When an electric current i is passed through the solenoid, a magnetic field is produced by inside is almost uniform and is directed along the axis of the solenoid as shown in Figure. The magnetic field at any point inside the solenoid is given by,
B = μ₀ni

As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines. The magnetic flux passing through each turn is
\({ \phi }_{B }=\int _{ A }^{ }{\vec B.d } \vec { A } =BAcos\theta =BA(since\theta=0^o\)
= (μ₀ni)A   ....(1)
The total magnetic flux linked or flux linkage of the solenoid with N turns (the total number of turns N is given by N= n l) is
\({ N\Phi }_{ B }\) = (nl)(μ₀ni)A
\({ N\Phi }_{ B }\) = (μ_on²Al)i    ...(2)
The equation (1) is,
\({ N\Phi }_{ B }\) = Li
Comparing equation (1) and (2), we have
L =μn²Al    ...(3)
From the above equation (3), it is clear that inductance depends on the geometry of the solenoid (turn density n, cross-sectional area A, length l) and the medium present inside the solenoid. If the solenoid is filled with a dielectric medium of relative permeability μ_r, then
L =μn²Al or L = μ₀ μ_r n² Al.

(i) Consider a straight conducting rod AB of length I in a uniform magnetic field \(\vec { B } \) which is directed perpendicularly into the plane of the paper.
(ii) The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec { v } \) towards right side
(iii) When the rod moves, the free electrons present in it also move with same velocity \(\vec { v } \) in \(\vec { B } \). As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation

\({ \vec { F } }_{ B }=-e(\vec v\times \vec { B } )\)
(iv) The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field \(\vec { E } \) directed along BA 
(v) Due to the electric field \(\vec { E } \), the coulomb force starts acting on the free electrons along AB and is given by
\({ \vec { F } }_{ E }=-e\vec { F } \)
(vi) The magnitude of the electric field \(\vec { E } \) keeps on increasing as long as accumulation of electrons at the end  A continues. The force \({ \vec { F } }_{ E }\) also increases until equilibrium is reached.
(vii) At equilibrium, the magnetic  Lorentz force \({ \vec { F } }_{ B }\) and the coulomb force \({ \vec { F } }_{ E }\) balance each other and no further accumulation of free electrons at the end A takes place.

\(\left| { \vec { F } }_{ B } \right| =\left| { \vec { F } }_{ E } \right| \) \(\left| -e(\vec { v } \times \vec { B } ) \right| =\left| -e\vec { E } \right| \) vB Sin 900 = E ⇒vB = E

The potential difference between two ends of the rod is
V = El
V = vBl
Thus, the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf
ε = Blv
As this emf is produced due to the movement of the rod, it is often called as motional emf. If the ends A and B are connected by an external circuit or total resistance R, then current \(i=\frac {ε}{R}=\frac{Blv}{R}\)flows in it. The direction of the current is found from right-hand thumb rule.

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

Capacitance of a parallel plate capacitor:
(i) Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d as shown in Figure.

(ii) The electric field between two infinite parallel plates is uniform and is given by \(E=\frac { \sigma }{ { \varepsilon }_{ o} } \) where σ is the surface charge density on the plates \(\left( \sigma =\frac { Q }{ A } \right) \).
iii) If the separation distance d is very much smaller than the size of the plate (d² < < A), then the above result is used even for finite-sized parallel plate capacitor.
The electric field between the plates is 
\(E=\frac { Q }{ A{ \varepsilon }_{ 0 } } ...(1)\)
iv) Since the electric field is uniform, the electric potential between the plates having separation d is given by
\(V=Ed=\frac { Qd }{ A{ \varepsilon }_{ 0 } } \quad \quad \quad ...(2)\)
Therefore the capacitance of the capacitor is given by
\(C=\frac { Q }{ V } =\frac { Q }{ \left( \frac { Qd }{ A{ \varepsilon }_{ 0 } } \right) } =\frac { { \varepsilon }_{ 0 }A }{ d } \quad \quad ....(3)\)
(v) From equation (3), it is evident that capacitance is directly proportional to the area of cross section and is inversely proportional to the distance between the plates.

Gauss law:
(i) A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in Figure. then the total electric flux through the closed surface of the sphere is
\(\Phi_E =\oint { \vec { E } .d\vec { A } =\oint { Ed } Acos\theta } \)     .....(1) 

(ii) The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element \(d\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0^o.
\(\\ \Phi_E =\oint { EdA } \)  Since cos0^o = 1   ......(2)
iii) E is uniform on the surface of the sphere, 
\(\\ \Phi_E=E\oint { dA } \)  .......(3)
Substituting for \(\oint { dA=4{ \pi r }^{ 2 } } \) and \(E=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { r }^{ 2 } } \) in eqn (3), we get
\(\therefore \phi E=4{ \pi r }^{ 2 }E\)
\({ \phi }_{ E }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { r }^{ 2 } } \times { 4\pi r }^{ 2 }=4\pi \frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } Q\)
\({ \phi }_{ E }=\frac { Q }{ { \varepsilon }_{ 0 } } \)   .....(4)
The equation (4) is called as Gauss's law.

Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q. This is shown in Figure.

Electrostatic potential at a point P
The electric potential at the point P is
\(V=\int _{ \infty }^{ r }{ \left( -\vec { E } \right) d\vec { r } } =-\int _{ \infty }^{ r }{ \vec { E } } .d\vec { r } \)  ...(1)
Electric field due to positive point charge q is
\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } \hat { r } \)
\(V=\frac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .d\vec { r } } \)
The infinitesimal displacement vector, \(d\vec { r } =dr\hat { r } \) and using \(\hat { r } \).\(\hat { r } \) = 1, we have
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .dr\hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } dr } } \)
After the integration,
\(V=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q{ \left\{ -\frac { 1 }{ r } \right\} }_{ \infty }^{ r }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)
Hence, the electric potential due to a point charge q at a distance r is
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)   ....(2) 

Torque experienced by an electric dipole in the uniform electric field:
Consider an electric dipole of dipole moment \(\vec { p } \) placed in a uniform electric field \(\vec { E } \) whose field lines are equally spaced and point in the same direction. The charge +q will experience a force  \(q\vec { E } \) in the direction of the field and charge -q will experience a force  \(-q\vec { E } \) in a direction opposite to the field. Since the external field \(\vec { E } \) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole.
The total torque on the dipole about the point O
\(\vec{\tau}=\overrightarrow{O A} \times(-q \vec{E})+\overrightarrow{O B} \times q \vec{E}\)         ......(1)
Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.

The magnitude of the total torque
\(\tau =|\vec { OA } |\left( -q\vec { E } \right) |sin\theta +|\vec { OB } ||\vec { E } |sin\theta \)
\(\tau =qE.2a\quad sin\theta \) ....(2)
where θ is the angle made by \(\vec { p } \) with \(\vec { E } \) since p = 2aq, the torque is written in terms of the vector product as 
\(\vec { \tau } =\vec { p } \times \vec { E } \)       ...(3)
The magnitude of this torque is \(\tau \) = pEsin \(\theta \)   and is maximum when θ = 90^o.
This torque tends to rotate the dipole and align it with the electric field \(\vec{E}\). Once \(\vec{p}\) is aligned with \(\vec{E}\), the total torque on the dipole becomes zero.