Yes, but motion is simple harmonic only when charge - q is not very far from the centre of ring on its axis. Otherwise motion is periodic, but not simple harmonic in nature.

Ex and by

The Ring Moves a way From the Solenoid

\(R = ρ L/A \)
(i.e) 10.25%

Power factor \(=\frac { R }{ Z } =\frac { R }{ \sqrt { { R }^{ 2 }+{ ({ X }_{ L }-{ X }_{ C }) }^{ 2 } } } \)
Max. value of power factor is 1, when Z = R
i.e., when \({ X }_{ L }={ X }_{ C },\) i.e ., at resonance, when the circuit is non-inductive.

(a) True 
\(\frac { h }{ e } =\frac { joule-sec }{ coulomb } =\frac { joule }{ ampere } \)
\( =magnetic\ flux\)
(b) False, for reason, see text.
(c) False. As magnetic flux does not change with time.
(d) true, as \(e=B\upsilon l\sin { \theta } .\)

A charged glass rod can lift a piece of paper against the gravitational pull of earth on this piece.

(i) Surface charge density,
\(\sigma=\frac{Q}{4 \pi R^2}\)
(ii) Potential at point P due to charge Q,
\(V_1=\frac{Q}{4 \pi \varepsilon_0(R / 2)}=\frac{2 Q}{4 \pi \varepsilon_0 R}\)
Due to spherical shell, potential at all point is same inside the shell and it is equal to potential at the centre of the shell.

Potential at P due to sphere, \(V_2=\frac{q}{4 \pi \varepsilon_0 R}\)
So, net potential at P,
\(\begin{aligned} V=\frac{2 Q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 R} \end{aligned}\)
\(\begin{aligned} V=\frac{12}{4 \pi \varepsilon_0}\left(\frac{2 Q}{R}+\frac{q}{R}\right) \end{aligned}\)

According to the Gauss's theorem, the total electric flux through a closed surface is \(\frac{1}{\varepsilon_{0}}\) times the magnitude of net charge enclosed by the surface. Consider a hollow charged sphere.
Take a point inside the sphere where electric field is to be calculated. Draw a Gaussian surface of radius r having point P on its surface.

Now, \(\phi=\oint_{S} \vec{E} \cdot \vec{d} s=E \oint_{S} d s=E .4 \pi r^{2}\)
According to the Gauss's theorem,
\(E .4 \pi r^{2}=\frac{q}{\varepsilon_{0}}(\because q=0)\)
\(\therefore E=0\)

If the field is non-uniform, the net force will be non-zero.

(i) Induced emf (voltage) in a coil \(\varepsilon\)= - L dI/dt
\(\frac{\varepsilon_1}{\varepsilon_2}=\frac{\mathrm{L}_1 \frac{d \mathrm{I}}{d t}}{\mathrm{~L}_2 \frac{d \mathrm{I}}{d t}}=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{4}{3}\)
(ii) Power supplied P = eI
As power is same for both coils
\(\begin{aligned} \varepsilon_1 \mathrm{I}_{\mathrm{I}} & =\varepsilon_2 \mathrm{I}_2 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{\mathrm{I}_1}{\mathrm{I}_2} & =\frac{\varepsilon_2}{\varepsilon_1}=\frac{3}{4} \end{aligned}\)
(iii) Energy stored in a coil \(\mathrm{U}=\frac{1}{2} \mathrm{LI}^2\)
\(\therefore \frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{\frac{1}{2} \mathrm{~L}_1 \mathrm{I}_1^2}{\frac{1}{2} \mathrm{~L}_2 \mathrm{I}_2^2}=\frac{\mathrm{L}_1 \mathrm{I}_1^2}{\mathrm{~L}_2 \mathrm{I}_2^2}=\frac{3}{4}\)

(i) Given, C₁ = 1\(\mu F\)  C₂ = 2\(\mu F\) C₃ = 3\(\mu F\)
The combined capacity (C_s) in series combination is given by
 \(\frac { 1 }{ { C }_{ s } } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } =\frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 11 }{ 6 } \)
\(\Rightarrow { C }_{ s }=\frac { 6 }{ 11 } \mu F\)
The combined capacity (C_p) in parallel combination is given by 
C_p = C₁ + C₂ +C₃ = 1 + 2 + 3 = 6\(\mu F\) 
\(\Rightarrow { C }_{ p }=11{ C }_{ s }\) 
(ii)  As, \({ \ C }_{ p }>{ C }_{ s }\) 
\(\therefore \) The charge supplied by 100 V battery
\({ q }_{ p }={ C }_{ p }V=6\mu F\times 100=6\times { 10 }^{ -6 }\times 100\)   
\(\\ { q }_{ p }=6\times { 10 }^{ -4 }C\)​​​​= 600 \(\mu\)C

\(Here,\ r=8cm=8\times { 10 }^{ -2 }m\)
\(\frac { dE }{ dt } =​​{ 10 }^{ 13 }V{ m }^{ -1 }{ s }^{ -1 }\)
Displacement current.
\({ I }_{ D }={ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } { { \epsilon }_{ 0 } }\frac { d }{ dt } (EA)\)
\( ={ \epsilon }_{ 0 }A\frac { { d }E }{ dt } ={ { \epsilon }_{ 0 }\pi { r }^{ 2 } }\frac { dE }{ dt } \)
\( =(8.85\times { 10 }^{ -12 })\times 3.14\times ({ 8\times { 10 }^{ -2 }) }^{ 2 }\times { 10 }^{ 13 }\)
\( =1.78\ A\)

Length of wire used, 
\(L=2\pi r\times no.\quad of\quad turns\)
\(=2\pi \times (2\times { 10 }^{ -2 })\times 500\quad m\)
\( Resistancperunitlength=0.02\Omega { m }^{ -1 }\)
\( Totalresistanceofwire,\)
\( R=2\pi \times (2\times { 10 }^{ -2 })\times 500\times .02=0.4\pi \Omega \)
\( No.of \ turn \ sperunit \ length,\)
\(n=\frac { 500 }{ 30\times { 10 }^{ -2 } } =\frac { 5000 }{ 3 } { m }^{ -1 }\)
\(AsB={ \mu }_{ o }nI={ \mu }_{ o }\frac { n\varepsilon }{ R }\)
\( So\varepsilon =\frac { BR }{ { \mu }_{ o }n } =\frac { { 10 }^{ -2 }\times \left( 0.4\pi \right) }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times \left( 5000/3 \right) } =6V\)

(a) Moving coil galvanometer is an instrument used for detection and measurement of small currents. Moving coil galvanometer works on the principle that when a current-carrying coil is placed in magnetic field it experiences a torque. 
(b) (i) Voltage sensitivity is defined as the deflection produced in the galvanometer when a unit voltage is applied across the two terminals of the galvanometer:
Vs = NBA/KR
(ii) Current sensitivity is defined as the deflection produced in the galvanometer when a limit current flow through the galvanometer:
Is = NBA/K
where N is number of turned in coil, B is the magnetic field, A is the area of the coil, K is the restoring torque per limit twist.

(a) Expression for the potential energy of a system of two point charges in an external field:
Work done in bringing the charge q₁from infinity to r₁ ,
Work done \(={ q }_{ 1 }V\left( { r }_{ 1 } \right) \)
Work done in bringing the charge qz from infinity to r₂ .
Work done against the external electric field
\(={ q }_{ 2 }V\left( { r }_{ 2 } \right) \)
Work done = work done against the external electric field + Work done on q₂ against the field due to q₁
\(={ q }_{ 2 }V\left( { r }_{ 2 } \right) +\frac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon  }_{ 0 }{ r }_{ 12 } } \)
Potential energy of the system = the total work done in assembling the configuration
\(={ q }_{ 1 }V\left( { r }_{ 1 } \right) +{ q }_{ 2 }V\left( { r }_{ 2 } \right) +\frac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ r }_{ 12 } } \)
(b) Electrostatic potential energy of the systems of charges corresponding to initial configuration is
\(U_{i}=\frac{2 k Q^{2}}{l}-\frac{3 k Q^{2}}{l}-\frac{6 k Q^{2}}{l}=-\frac{7 k Q^{2}}{l}\)
Electrostatic potential energy of the system of charges corresponding to final configuration is
\(U_{f}=\frac{4 k Q^{2}}{l}-\frac{6 k Q^{2}}{l}-\frac{12 k Q^{2}}{l}=-\frac{14 k Q^{2}}{l}\)
The amount of work done in shifting the charges to new locations is
\(W=U_{f}-U_{i}\)
\(=-\frac{14 k Q^{2}}{l}-\left(-7 \frac{k Q^{2}}{l}\right)=-\frac{7 k Q^{2}}{l}\)

(a) Electric field intensity due to infinitely long charged straight wire: Consider a linear charge distribution with charge density A. We imagine a symmetrical Gaussian surface around length I of this distribution in such a way that the point P where we have to calculate electric field lies on it.

Electric flux through the circular faces of this Gaussian surface is zero.
\( \phi_{s}=\int \vec{E} \cdot \overrightarrow{d s}=\text { E.ds } \cos 90^{\circ}=0 \left(\because \theta=90^{\circ}\right) \)
Electric flux through the curved surface is given by
\( \phi_{c s}=\oint \vec{E} \cdot \overrightarrow{d s}=\oint E \cdot d s \)
\(\phi_{c s}=E \oint d s=E(2 \pi r l) \ \left(\because \theta=0^{\circ}\right) \)
Net flux through the Gaussian surface is given by
\(\phi_{E}=\phi_{s}+\phi_{c s}=E(2 \pi r l)...(i)\)
According to the Gauss's theorem
\(\phi_{E}=\frac{q}{\varepsilon_{0}}=\frac{\lambda l}{\varepsilon_{0}} \ (\because q=\lambda l) \ \ldots(i i)\)
From equations (i) and (ii), we get
\(E=\frac{\lambda}{2 \pi \varepsilon_{0} r} \Rightarrow E \propto \frac{1}{r}\)
(b) 

(c) Work done in moving a charge 'q' through a small displacement \(\overrightarrow{d r}\) is given by
\( d W =\vec{F} \cdot d \vec{r}=q \vec{E} \cdot \overrightarrow{d r} \)
\(d W =q E d r \cos \theta\left(\theta=0^{\circ} \text { and } \cos 0^{\circ}=1\right) \)
\(d W =q \times \frac{\lambda}{2 \pi \varepsilon_{0} r} d r \)
\(E =\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
Work done in moving the given charge from rl to \(r_{2}\left(r_{2}>r_{1}\right)\)
\( W=\int_{r_{1}}^{r_{2}} d W=\frac{\lambda q}{2 \pi \varepsilon_{0}} \int_{r_{1}}^{r_{2}} \frac{d r}{r} \)
\(W=\frac{\lambda q}{2 \pi \varepsilon_{0}}\left[\log _{e} r_{2}-\log _{e} r_{1}\right] \)
\(W=\frac{\lambda q}{2 \pi \varepsilon_{0}} \log _{e} \frac{r_{2}}{r_{1}} \)

Current flowing through various branches of the circuit is represented in the given figure.

I₁ = Current flowing through the outer circuit
I₂ = Current flowing through branch AB
I₃ = Current flowing through branch AD
I₂ - I₄ = Current flowing through branch BC
I₃ + I₄ = Current flowing through branch CD
I₄ = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I₂ + 5I₄ - 5I₃ = 0
2I₂ + I₄ - I₃ = 0
I₃ = 2I₂ + I₄ … (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I₂ - I₄) - 10(I₃ + I₄) - 5I₄ = 0
5I₂ + 5I_{4 -} 10I₃ - 10I₄ - 5I₄ = 0
5I₂ - 10I₃ - 20I₄ = 0
I₂ = 2I₃ + 4I₄ … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
-10 + 10 (I₁) + 10(I₂) + 5(I₂ - I₄) = 0
10 = 15I₂ + 10I₁ - 5I₄
3I₂ + 2I₁ - I₄ = 2 … (3)
From equations (1) and (2), we obtain
I₃ = 2(2I₃ + 4I₄) + I₄
I₃ = 4I₃ + 8I₄ + I4
- 3I₃ = 9I₄
- 3I₄ = + I₃ … (4)
Putting equation (4) in equation (1), we obtain
I₃ = 2I₂ + I₄
- 4I₄ = 2I₂
I₂ = - 2I₄ … (5)
It is evident from the given figure that,
I₁ = I₃ + I₂ … (6)
Putting equation (6) in equation (1), we obtain
3I₂ +2(I₃ + I₂) - I₄ = 2
5I₂ + 2I₃ - I₄ = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(- 2 I₄) + 2(- 3 I₄) - I₄ = 2
- 10I₄ - 6I₄ - I₄ = 2
17I₄ = - 2
\(I_{4}=-\frac{2}{17} A\)
Equation (4) reduces to
I₃ = - 3(I₄)
\(=-3\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{2}=-2\left(I_{4}\right)\)
\(=-2\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{2}-I_{4}=\frac{4}{17}-\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{3}+I_{4}=\frac{6}{17}+\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{1}=I_{3}+I_{2}\)
\(=\frac{6}{17}+\frac{4}{17}=\frac{10}{17} A\)
Therefore, current in branch \(A B=\frac{4}{17} A\)
In branch BC \(=\frac{6}{17} A\)
In branch CD = \(-\frac{4}{17} A\)
In branch AD = \(\frac{6}{17} A\)
In branch BD = \(\left(-\frac{2}{17}\right) A\)
Total current = \(\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} A\)

Principle: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it (mutual induction).
Derivation :
The induced emf or voltages Es. in the secondary, with Ns, turns, is
\({ E }_{ s }=\frac { -{ N }_{ s }d\phi }{ dt } \)
The alternating flux also induces an emf, called back emf, in the primary. This is
\({ \varepsilon }_{ p }=\frac { -{ N }_{ p }d\phi }{ dt } \)
But
\({ \varepsilon }_{ s }={ V }_{ s }>{ \varepsilon }_{ p }={ V }_{ p }\)
therefore,
\({ V }_{ s }=\frac { -{ N }_{ s }d\phi }{ dt } \quad and\quad { V }_{ p }=\frac { -{ N }_{ p }d\phi }{ dt } \)
Hence
\(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ { N }_{ p } } \)
If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output,
\(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ p } }{ { I }_{ s } } \)
The large-scale transmission and distribution of electrical energy over long distances are done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the 12 R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.

Given,  
\(L=80mH=80\times { 10 }^{ -3 }H, \ R=0, \ v=50Hz\)
\(C=60\mu F=60\times { 10 }^{ -6 }F,\)
\(\omega =2\pi v=100\pi \ rad/s \)
\({ V }_{ rms }=230 \ V,\)
and \(\\ { V }_{ 0 }=\sqrt { 2{ V }_{ rms } } =\sqrt { 2 } \times 230V\)
(i) I₀ = ? and I_rms = ?
\(\begin{aligned} \Rightarrow I_0 & =\frac{V_0}{Z}=\frac{V_0}{\left|\omega L-\frac{1}{\omega C}\right|} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{230 \sqrt{2}}{\left|100 \pi \times 80 \times 10^{-3}-\frac{1}{100 \pi \times 60 \times 10^{-6}}\right|} \\ \end{aligned}\)
\(\begin{aligned} =\frac{230 \sqrt{2}}{\left|8 \pi-\frac{1000}{6 \pi}\right|}=-11.63 \mathrm{~A} \\ \end{aligned}\)
\(\begin{aligned} I_{\mathrm{rms}} & =\frac{I_0}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}=8.23 \mathrm{~A} \end{aligned}\)
(ii) For L, V_L = I_rms \(\omega \)L = 8.23 \(\times\) 100\(\pi\)\(\times\)80 \(\times\)10^-3
= 206.84 V
For C, \(V_C=I_{\mathrm{rms}} \frac{1}{\omega C}=8.23 \times \frac{1}{100 \pi \times 60 \times 10^{-6}}\)
= 436.84 V
Since, voltage across L and C are 180° out of phase, therefore they are subtracted.
Thus, applied rms voltage = 436.84 - 206.84
= 230.0 V
(iii) Average power transferred per cycle by source to inductor is always zero because of phase difference of \(\pi\)/2 between voltage and current through inductor.
(iv) Average power transferred per cycle by source to inductor is always zero because of phase difference of \(\pi\)/2 between voltage and current through inductor
(v) \(\therefore\) Total average power absorbed by the circuit is also zero.

There are two protons p₁ and p₂ with an electron e.
Distance between two protons id given by
r₁ = 1.5\(\overset { o }{ A } \) = 1.5 x 10^-10 m

Distance between proton p1 and electron e is given by
r2 = 1 \(\overset { o }{ A } \)
 = 1 x 10-10 m
Distance between proton p2 and electron e is given by
r3 = 1\(\overset { o }{ A } \)
=1 x10-10 m
The total potential energy of the system,
\(U=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\left[ \frac { { q }_{ p1 }{ q }_{ p2 } }{ { r }_{ 1 } } +\frac { { q }_{ p1 }{ q }_{ e } }{ { r }_{ 2 } } +\frac { { q }_{ p2 }{ q }_{ e } }{ { r }_{ 13 } } \right] \)......(i)
Given qp1 = qp2
= 1.6 x10-19 C
and  qe= -1.6 x10-19 C
Putting these values in Eq.(i), we get
\(U=9 \times 10^{9}\left[\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{1.5 \times 10^{-10} \cdots}\right.\) \(+\frac{\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{10^{-10}}\) \(\left.+\frac{1.6 \times 10^{-19} \times\left(-1.6 \times 10^{-19}\right)}{10^{-10}}\right]\) 
\(=\frac{9 \times 10^{9} \times 1.6 \times 1.6 \times 10^{-38}}{10^{-10}}\left[\frac{1}{1.5}-1-1\right]\)
= -30.72 x 10-19 J
\(=\frac{-30.72 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=-19.2 \mathrm{eV}\)
Here, we use that potential energy at infinity is zero.

Here A = 1m², d = 5 cm, q =?
E = 1000 NC^-1
From \(E={\sigma\over\epsilon_o}={q/A\over \epsilon_o}\)
\(q=A\epsilon_oE=1\times (8.85\times 10^{-12})\times 1000\)
\(=8.85\times 10^{-9}C\)