Let us consider a circular loop of radius a with centre C. Let the plane of the coil be
perpendicular to the plane of the paper and current I be flowing in the direction as shown in the figure. Suppose P is any point on the axis at a direction from the centre.

Now, consider a current element Idl on top (L) where current comes out of paper normally, whereas at bottom (M) enters into the plane of paper normally.
\(\because LP \bot Idl\)
Also \(MP \bot Idl\)
\(\because LP=MP=\sqrt{r^{2}+a^{2}}\)
The magnetic field at point P due to current element Idl. According to Biot-Savart's law,
dB = \(\frac{\mu_0}{4\pi}.\frac{Idl sin 90^{0}}{(r^{2}+a^{2})}\)
where, a = radius of circular loop,
r = distance of point P from centre along the axis. The direction of dB is perpendicular to LP and along PQ, where \(PQ \bot LP\)  Similarly, the same magnitude of magnetic field is obtained due to current element Idl at the bottom and direction is along PQ', where \(PQ^{'} \bot MP\).
Now, resolving dB due to current clement at Land M dB cos \(\phi\) components balance each other and net magnetic field is given by
B = \(\oint dB sin \phi = \oint \frac{\mu_0 }{4\pi} (\frac{Idl}{r{2}+a^{2}}).\frac{a}{\sqrt{r^{2}+a^{2}}}\)
\([\therefore In \Delta PCL, sin \phi = \frac{a}{\sqrt{r^{2}+a^{2}}}]\)
\(=\frac{\mu_0}{4\pi} \frac{Ia}{(r^{2}+a^{2})^{3/2}} \oint dl=\frac{\mu_0}{4\pi}\frac{Ia}{(r^{2}+a^{2})^{3/2}}(2\pi a)\)
or  \(B=\frac{\mu_0Ia^{2}}{2(r^{2}+a^{2})^{3/2}}\)
For N turns, B = \(\frac{\mu_0 N Ia^{2}}{2(r^{2}+a^{2})^{3/2}}\) Tesla.
The diagram of magnetic field lines due to a circular wire carrying current I is.

Here, \(L=200mH=200\times 10^{ -3 }H\) , \(E_{ 0 }=210\ V\),
\(v=50\ Hz,\ I_{ 0 }=?\)
\(I_{ 0 }=\frac { E_{ 0 } }{ { X }_{ L } } =\frac { E_{ 0 } }{ \omega L } =\frac { E_{ 0 } }{ 2\pi vL } =\frac { 210 }{ 2\times 3.14\times 50\times 0.2 } \)
\(=3.3 \ A\)
Through L, current lags behind the voltage by a phase angle of 90^o. Therefore, when current is at its peak value, voltage must be zero.

Here \(\overrightarrow { E } =5\times 10^{ 3 }\hat { i } NC^{ -1 }\)
A = (10 cm)² = (10^-1m)² = 10^-2m²
When plane is || to Y-Z plane, \(\theta =0^o\)
\(\phi=EA \ cos \theta=(5\times 10^3)\times 10^{-2}cos \ 0^o\)
= 50 Nm²C^-1
When plane makes an angle of 30^o with X-axis,
\(\theta =60^o\)
\(\phi'=EA \ cos\theta=5\times 10^3\times10^{-2}cos \ 60^o\)
\(=25 Nm^2C^{-1}\) 

Total required capacitance, C = 2 µF
Potential difference, V = 1 kV = 1000 V
Capacitance of each capacitor, C₁ = 1 µF
Each capacitor can withstand a potential difference, V₁ = 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as \(\frac{1000}{400}=2.5\)
Hence, there are three capacitors in each row.
Capacitance of each row \(=\frac{1}{1+1+1}=\frac{1}{3} \mu \mathrm{F}\)
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
\(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\ldots \ldots \ldots \ldots \ldots \ldots . n \text { terms }\)
\(=\frac{n}{3}\)
However, capacitance of the circuit  is given as 2 \(\mu \) F
\(\therefore \frac{n}{3}=2\)
n = 6
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 x 3 i.e., 18 capacitors are required for the given arrangement.

If I is the current in the circuit containing inductor of inductance L, capacitor of capacitance C and resistor of resistance R in series, then the voltage drop across the inductor is

which leads current I by phase angle of \(\pi/2\), and voltage drop across the capacitor is V₂ = 1 X X_C. which lags behind current I by phase angle of \(\pi/2\) and voltage drop across the resistor is V_R = IR which is in phase with current I. So the net voltage E across the circuit is (using phasor diagram)

E = \(\sqrt{{V }_{ R }^{ 2 }+{({V}_{L}-{V}_{C})}^{2}}\)
\(\Rightarrow\) \(E=I\sqrt{{R}^{2}+{({X}_{L}-{X}_{C})}^{2}}\Rightarrow E=IZ\)
where,Z = \(\sqrt{{R}^{2}+{({X}_{L}-{X}_{C})}^{2}}\) is known as impedance, Phase angle between voltage and current is given by \(tan\ \phi={{{V}_{L}-{V}_{C}}\over{{V}_{R}}}={{{X}_{L}-{X}_{C}}\over{{R}}}.\) A series L-C-R circuit its natural angular frequency, \(\omega={{1}\over{\sqrt{LC}}}\) and natural ( resonating ) frequency, v = \({{1}\over{2\pi\sqrt{LC}}}\)
when the applied AC in the circuit has this frequency the series L-C-R circuit offers minimum impedance i.e. only R and current at this frequency flows maximum. In the case of resonance, voltage and current are in same phase. Above mentioned condition is known as condition of resonance. In this condition (1)
Inductive and capacitive reactances are equal
\({X}_{L}={X}_{C}\Rightarrow\omega L={{1}\over{\omega C}}\Rightarrow \omega={{1}\over{\sqrt{LC}}}\)  \([\because \ \omega=2\pi v]\)
\(\Rightarrow\)  v = \({ { 1 }\over{ 2\pi\sqrt{LC} } }\)
Potential. drop across inductor and capacitor are equal. V_L = V_C
The series resonant circuit is also called an acceptor circuit because when a number of different frequency currents are into the circuit, the circuit offers minimum impedance to natural frequency current.
For L-R circuit, X_L = R
Power factor,P₁= cos\(\phi\)
= \({{R}\over{\sqrt{{R}^{2}+{({X}_{L}-{X}_{C})}^{2}}}}={{R}\over{\sqrt{{R}^{2}+{({X}_{L}-{X}_{L})}^{2}}}}={{R}\over{R}}=1\)
Required ratio = \({{{P}_{1}}\over{{P}_{2}}}={{1}\over{\sqrt{2}}}\)

Here, v=2.18 x 10⁶ ms^-1,
r=5.3 x 10^-11 m, e=1.6 x 10^-19 C.
(a) Time period of revolution of electron is given by, 
\(T=\frac { 2\pi r }{ v } =\frac { 2\pi \times 5.3\times { 10 }^{ -11 } }{ 2.18\times { 10 }^{ 6 } } =1.528\times { 10 }^{ -16 }s\) 
Equivalent current, \(I=\frac { charge }{ time } =\frac { e }{ T } \) 
\(=\frac { 1.6\times { 10 }^{ -19 } }{ 1.528\times { 10 }^{ -16 } } =1.05\times { 10 }^{ -3 }A\) 
\((b)B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi I }{ r } =\frac { { 10 }^{ -7 }\times 2\pi \times 1.05\times { 10 }^{ -3 } }{ 5.3\times { 10 }^{ -11 } } \) 
=12.4 T

(i) The rays coming from fire are yellow, red and infrared rays
(ii) The infrared rays coming from fire provide soothing effect to the body muscles of the sick person sitting near the fire, Due to it, the sick person feel better after some time
(iii) From the above study we learn that presence of mind and proper use of the things available help ton save a difficult situation

Here, I_v = ? I₀ = 5.0 A
As \({ I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } \)
\(\therefore \ \ { I }_{ v }=\frac { 5 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 5\times 1.414 }{ 2 } =3.54A\)
(i) An a.c. ammeter will read 3.54 ampere
(ii) An ordinary moving coil ammeter will read zero because it records average value of current over a complete cycle, which is zero in case of alternating current.

Here, I₀ = 10A, R = 20\(\Omega \), L = 10H
(i) I = ?, t = 0.4 sec.
During decay, I = I₀ e^{-(R/L) t} = 10e^-20\(\times\)^0.4/10 = 10e^-0.8
log₁₀ I = log₁₀10 - 0.8 log₁₀e
= 1-0.8\(\times\)log₁₀ 2.718 = 1-0.8\(\times\)0.4343
 log₁₀ I = 1-0.3474 = 0.6526
I = antilog 0.6526 = 4.49A
(ii) t = ? I =\(\frac { 60 }{ 100 } \) I₀ From I = I₀ e^{(-R/L) t}
 \(\frac { I }{ { I }_{ 0 } } ={ e }^{ (-R/L)t }\ ;\ \frac { 60 }{ 100 } \ { e }^{ \frac { -20 }{ 10 } t }\ =\ { e }^{ -2t }\)
log₁₀ 6 - log₁₀10 = -2t log₁₀e
 0.7782 - 1.0000 = 1.0000 = -2 t\(\times\) 0.4343
or -0.2218 = -t\(\times\)0.8686
t = \(\frac { 2.2218 }{ 0.8686 } \) = 0.255s

Here, R = 10 cm = 0.10 cm;
C = 200 pF = 200 x 10^-12 F, 
\(\omega\)=400 rad s^-1 , V_rms = 230 V.
(i) \({ I }_{ rms }=\frac { { V }_{ rms } }{ { X }_{ C } } =\frac { { V }_{ rms } }{ 1/\omega C } ={ V }_{ rms }\times \omega C\)
= 230 x 400 x (200 x 10^-12)
= 18.4 x 10^-6 A = 18.4 \(\mu A\)
(ii) Magnetic field at a distance r from the axis of plates is
\(B=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } I\)   [See Solved Example 5]
Amplitude of \(\overrightarrow { B } \) is given by
\({ B }_{ 0 }=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } { I }_{ 0 }=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } { I }_{ rms }\sqrt { 2 } \)
\(\left[ \because { I }_{ rms }={ I }_{ 0 }/\sqrt { 2 } \right] \)
Here, r = 2.0 cm = 2.0 x 10^-2 m ; R = 0.10 m.
\(\therefore { B }_{ 0 }=\frac { \left( 4\pi \times { 10 }^{ -7 } \right) }{ 2\pi } \times \frac { 2.0\times { 10 }^{ -2 } }{ { \left( 0.10 \right) }^{ 2 } } \times \left( 18.4\times { 10 }^{ -6 } \right) \)
= 1.04 x 10^-11

Useful Intensity,
\(I=\frac { power }{ area } =\frac { 100\times \left( 2.5/100 \right) }{ 4\pi { \left( 3 \right) }^{ 2 } } =\frac { 2.5 }{ 36\pi } W{ m }^{ -2 }\)
Half of this intensity (I) belongs to electric field and half of that to magnetic field. Therefore,
\(\frac { I }{ 2 } =\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c \ or \ { E }_{ 0 }=\sqrt { \frac { 2I }{ { \varepsilon }_{ 0 }c } } \)
\(=\sqrt { \frac { 2\times \left( 2.5/36\pi \right) }{ \left( \frac { 1 }{ 4\pi \times 9\times { 10 }^{ 9 } } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } =4.08V{ m }^{ -1 }\)
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.08 }{ 3\times { 10 }^{ 8 } } =1.36\times { 10 }^{ -8 }T\)

Power is the rate of doing work. In A.C current and voltage change at every moment. Since power is the product of instantaneous voltage and current, therefore it changes at every moment.
To determine average power in a circuit containing pure resistor, determine total energy spent in one cycle and divide it by the period.
Instantaneous power  \(=EI=\left[ { E }_{ 0 }sin\omega t \right] \left[ { I }_{ 0 }sin\omega t \right] \)
Instantaneous work done in small interval dt is
\( \ dW=EI \ dt={ E }_{ 0 }{ I }_{ 0 }{ sin }^{ 2 }\omega t \ dt\)
\(W=\int _{ 0 }^{ T }{ { E }_{ 0 }{ I }_{ 0 }{ sin }^{ 2 }\omega t \ dt }\)
\(W={ E }_{ 0 }{ I }_{ 0 }\int _{ 0 }^{ T }{ { \left( \frac { 1-cos2\omega t }{ 2 } \right) }dt } \)
\(=\left[ \because \int _{ 0 }^{ T }{ dt } -\int _{ 0 }^{ T }{ cos2\omega t } \right]\)
\(or \ \ W=\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } \left[ T-0 \right] =\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } T\)
\( \left[ \because \int _{ 0 }^{ T }{ dt } -\int _{ 0 }^{ T }{ cos2\omega t } \right] =0\)
\(or \ \ { P }_{ av }=\frac { W }{ T } =\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } =\frac { { E }_{ 0 } }{ \sqrt { 2 } } .\frac { { I }_{ 0 } }{ \sqrt { 2 } } \)
\(\ {P }_{ av }={ E }_{ v }{ I }_{ v }\)
\(\therefore\) Average power over a complete cycle of a.c through a resistor is the product of virtual voltage and virtual current.

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

Let us consider a charge q is placed at the centre of a cubic Gaussian surface. As per the question,
q = 2  \(\mu\)C = 2 \(\times\) 10^-6 C
Length of edge of = 9 cm  ....(1)
 
According to Gauss' theorem, the net electric flux \((\phi)\) through the surface is given by
\(\phi=\frac { q }{ { \varepsilon }_{ 0 } }=\frac { 2\times { 10 }^{ -6 } }{ 8.854\times { 10 }^{ -12 } }=2.26\times{ 10 }^{ 5 }\) N - m² /c
Thus, the net electric flux through the surface is 2.26 \(\times\) 10⁵ N-m² /C.

One ampere of current can be defined as the amount of current which when flows through two infinitely long parellel wires seperated by one metere produces an attractive foce 2 x 10^-7N/m.

Let R₀ be the resistance of the conductor at 0^oC. Then resistance of conductor B at 0^oC = n R₀.
Resistance of conductor A at \(\theta^oC\),
R₁ = R₀(1 + \(\alpha_1 \theta\))
Resistance of conductor A at  \(\theta^oC\) ,
R₂ = n R₀(1 + \(\alpha_2 \theta\))
Total resistance,
R_s = R₁ + R₂ 
= R₀(1 + \(\alpha_1 \theta\)) +  n R₀(1 + \(\alpha_2 \theta\))
= \((1+n)R_0[1+\frac{\alpha_1+n \alpha_2}{1+n}\theta]\)
Comparing this relation with
R_s = R_s0 [ 1 + \(\alpha_s \theta\))
We have, resistance of series combination at \(\theta^oC\) 
R_s0 = (1 + n)R₀
Temperature coefficient of resistance of the series combination is
\(\alpha_s=\frac{\alpha_1+n\alpha_2}{1+n}\) 

A choke coil reduces the voltage across the fluorescent tube without wastage of power.

To start a car, a high current is required which cannot be obtained from the series combination of 8 simple cells, because their internal resistance is of the order of \(10 Ω \) while the resistance of the car battery is only of the order of \(0.1 Ω.\)