∴ Displacement current,
I_d = \(\varepsilon _{ 0 }\frac { d\phi _{ E } }{ dt } \) = \(\varepsilon _{ 0 }\frac { d }{ dt } \) (EA)    \(\left[\because \phi_E=E A\right]\)
where, E is electric field and A is the area of cross-section
= \(\varepsilon _{ 0 }A\frac { d }{ dt } \)\(\left( \frac { V }{ d } \right) \)
\(\Rightarrow \) I_d =\(\frac { \varepsilon _{ 0 }A }{ d } \)x\(\frac { dV }{ dt } \)​​= \(\frac { CdV }{ dt } \) \(\left[\because C=\frac{\varepsilon_{0} A}{d}\right]\)​​​
\(\Rightarrow \)  \(\frac { dV }{ dt } \) = \(\frac { I_{ d } }{ C } \) = \(\frac { 1.0 }{ 10^{ -6 } } \) 
= 10⁶ Vs^-1
Thus, an instantaneous displacement current of I A can be set up changing the potentil difference across the paralled plates of capcitor at the rate of 10⁶ Vs^-1..

(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

Given = 900 = 3 x 10^-4 N-m
and B = 0.4 x 10^-4T
Since, torque is given by
\( \tau = MB\sin { \theta } \)
 \(\Rightarrow \ M\ =\frac { 3\times { 10 }^{ -4 } }{ 0.4\times { 10 }^{ -4 }\sin { \theta } } \)
\(\Rightarrow M\ =\ 7.5\ { jT }^{ -1 }\)

Given, resistance, \({ R }_{ 1 }=10\Omega ,\ { R }_{ 2 }=10\Omega \)
Voltage, V = 1V
Time t = 5 x 60 s = 300 s
Since, effective resistance in parallel combination will be
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } =\frac { 10\times 10 }{ 10+10 } =5\Omega \)        
Heat produced = \(\frac { { v }^{ 2 }t }{ { R }_{ p } } =\frac { { 1 }^{ 2 } }{ 5 } \times 5\times 60=\frac { 60 }{ 4.2 } cal=14.3\ \)cal

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

When we start the motor, there is no back emf as the motor is at rest. So, a large current flows through the coil. As the motor rotates, the back emf increases and intake of current decreases.

Asha displayed the values of awareness towards energy saving and her concern towards wastage of energy
Following measures should be adopted to minimize the wastage of electric energy in Our households:
We should switch off the electrical appliances which are not in use.
The old electrical appliances should be repaired time to time, as they often consume more electricity.
Condensers of electrical appliances should be replaced time to time.
New electrical and electronic appliances must be purchased in accordance with their efficiency to consume less energy.
Use of solar devices like solar heater and cookers must be encouraged rather than the use of conventional electrical appliances.
Electrical energy dissipated by an appliance is
given by E = P x t
where, P is the power and t is the time.
P = 2 W, t = 20 h = 20 X 60 X 60 s 
So, E = 2 x 20 x 60 x 60
\(\Rightarrow\)  E =1.44 x 10⁵ J

(a) Presence of mind, knowledge of subject.
(b) Used in telecommunication.
(c) \(\lambda =e/v=(3\times 10^{ 8 })/(3\times 10^{ 11 })={ 10 }^{ -3 }\)

(a) \(\overrightarrow{F}=I(\overrightarrow{l} \times \overrightarrow{B})\), where \(\overrightarrow{l}\) is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. According to Fleming's left hand rule,\(\overrightarrow{B}\) must act horizontally in a direction perpendicular to the wire carrying current.
(b) Adaptation to different situations and flexible and adjustable attitude.
(c) Avoiding unnecessary arguments in conflicting situations in everyday life.

(i) Lat carbon resistor S is given to the bridge. Then,
\(\frac { 2R }{ R } =\frac { 2R }{ S } \Rightarrow \frac { R }{ S } =1\)
R = S = 22 x 10³ \(\Omega =22k\Omega \)
(ii) After interchanging the resistance, the balanced bridge would be
\(\frac { 2R }{ X } =\frac { 22\times { 10 }^{ 3 } }{ 2\times 22\times { 10 }^{ 3 } } =\frac { 1 }{ 2 }\)
X = 4R = 4 x 22 x 10³
= 88 x 10³ \(\Omega \)
The colour sequence of X is grey, grey and orange. Thus, equivalent resistance of Wheatstone bridge,
\(\frac { 1 }{ { R }_{ eq } } =\frac { 1 }{ 3R } +\frac { 1 }{ 6R } =\frac { 3 }{ 6R } \)     
\( { R }_{ eq }=2R\)
\(\therefore \ Current \ through \ the \ circuit,I=\frac { 1 }{ 3 } \times \frac { V }{ 2R } =\frac { V }{ 6R } A\)

Given, number of turns N = 120, current I = 3 A, radius of coil, r = 18 cm = 0.18 m and distance from the centre to a point on axis,
a = r = 0.18 m
As, \(\begin{aligned} B=\frac{\mu_0 N I a^2}{2\left(a^2+r^2\right)^{3 / 2}} \\ \end{aligned}\)
\(\begin{aligned} =\frac{4 \pi \times 10^{-7} \times 120 \times 3 \times(0.18)^2}{2\left[(0.18)^2+(0.18)^2\right]^{3 / 2}} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad B & =4.4 \times 10^{-4} \mathrm{~T} \end{aligned}\)

\(Here, \ { E }_{ v }=200V,\ v=50hz,\ R=100\Omega ,\ { V }_{ R }=65V,\ { V }_{ C }=415V,\ { V }_{ L }=204V\)
(i) If Iv is current in the circuit, then
\({ V }_{ R }={ I }_{ V }\times R; \ 65={ I }_{ V }\times 100, \ { I }_{ V }=0.65A\)
\( (ii) \ { V }_{ L }={ I }_{ V }{ X }_{ L }; \ { X }_{ L }=\frac { { V }_{ L } }{ { I }_{ v } } =\frac { 204 }{ 0.65 } =313.85\Omega\)
\({ X }_{ L }=\omega L=2\pi vL=313.85\)
\(L=\frac { 313.85 }{ 2\pi v } =\frac { 313.85 }{ 3.14\times 50 } =1.0H\)
\( (iii) \ { V }_{ C }={ I }_{ v }{ X }_{ C }, \ { X }_{ C }=\frac { { V }_{ C } }{ { I }_{ v } } =\frac { 415 }{ 0.65 } =638.5\Omega \)
\({ X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } ; \ C=\frac { 1 }{ 2\pi v \ { X }_{ c } } =\frac { 1 }{ 2\times 3.14\times 50\times 638.5 } =4.99\times { 10 }^{ -6 }F\)
(iv) Let C′ be the capacitance that would produce resonance with L= 1.0H, then
\( v=\frac { 1 }{ 2\pi \sqrt { LC' } } ; \ C'=\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 }L } =\frac { 1 }{ 4\times \left( 3.14 \right) ^{ 2 }\times \left( 50 \right) ^{ 2 }\times 1 } =10.1\times { }^{ }F=10.1\mu F\)

Here, inductance, L = 2 henry
r.m.s. voltage, E_V = 150 volt
frequency of A.C. supply, v = 50 c/s.
\(\therefore\) Inductive reactance, X_L = \(\omega\)L
\(=2\pi vL=2\times \frac { 22 }{ 7 } \times 50\times 2=\frac { 4400 }{ 7 } \ ohm\)
If E₀ is the peak value of the alternating voltage, then maximum value of current (I₀) is given by
\({ I }_{ 0 }=\frac { { E }_{ 0 } }{ { X }_{ L } } \ or \ { I }_{ 0 }=\frac { \sqrt { 2 } \times { E }_{ v } }{ { X }_{ L } } =\frac { \sqrt { 2 } \times 150 }{ 4400/7 } =0.337A\)

Here, \(r=6\times { 10 }^{ -2 }m;\)
\(A=\pi \times \left( 6\times { 10 }^{ -2 } \right) ^{ 2 }=36\pi \times { 10 }^{ -4 }{ m }^{ 2 }\)
\(\frac { dE }{ dt } ={ 10 }^{ 10 }V{ cm }^{ -1 }{ s }^{ -1 }={ 10 }^{ 12 }V{ m }^{ -1 }{ s }^{ -1 }\)
Displacement current,
\({ I }_{ D }={ \epsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }A\frac { dE }{ dt } \)
\(=\left( 8.85\times { 10 }^{ -12 } \right) \times \left( 36\pi \times { 10 }^{ -4 } \right) \times { 10 }^{ 12 }\)
= 0.1 A

Quality factor
The quality factor of LCR-circuit is the ratio of the potential difference across inductance (or capacitance) at resonance to the applied voltage. This is also called the figure of merit or Q-value of LCR series circuit
i.e.   \(Q=\frac { Voltage \ across(L\ or\ C) }{ Voltage\ applied } \)
At resonance
Voltage across \(L=I{ X }_{ L }\)
Voltage across \(C=I{ X }_{ C }\)
Voltage applied = Voltage drop across \(R=IR\)

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

In case of metallic sphere, charge given to it mostly resides on its surface. Therefore, there is no difference whether the sphere is hollow or solid. As in both the cases, the charge that will reside will be same.

Let q be the charge on the charged capacitor.
\(\therefore \)  Energy stored in it is given by, \(U=\frac { { q }^{ 2 } }{ 2C } \) 
When another uncharged similar capacitor is connected, then the net capacitance of the system is given by, C' = 2C
The charge on the system remains constant.
So, the energy stored in the system is given by
\(U'=\frac { { q }^{ 2 } }{ 2C' } =\frac { { q }^{ 2 } }{ 4C } \)  \(\left[ \because { C'={ 2C } } \right] \)  
Thus, the required ratio is given by \({ \frac { U' }{ U } }\ =\frac { { { q }^{ 2 }/{ 4C } } }{ { q }^{ 2 }/{ 2C } } =\frac { 1 }{ 2 } \)

We know that resistivity of iron is more than that of copper.
In series combination, the same current I flows through iron and copper wires.
Heat produced,
H = I²Rt
= \(I^2\frac{\rho l}{A}\) i.e.,  \(H \propto \rho\)
As \(\rho_{iron}>\rho_{cu}\)
so \(H_{iron}>H_{cu}\).
Therefore, iron will start glowing first in series combination. In parallel combination, the voltage V is same across iron and copper wires. So
Heat produced,
\(H=\frac{V^2}{R}t=\frac{V^2t}{\rho l/A}\) i.e., \(H \propto \frac{1}{\rho}\)
As \(\rho_{iron}>\rho_{cu}\)
Therefore, copper will start glowing first in parallel combination of wires.

It is possible only if the fall of potential across the potentiometer wire is less than the potential difference to be balanced by the potentiometer wire.

Induced current,
\(I=\frac { induced \ e.m.f/ }{ resistance } =\frac { e }{ R } \\ =\frac { d\Phi /dt }{ R } =\frac { -1 }{ R } \frac { d }{ dt } (BA \ cos{ 0 }^{ \circ })\\ I=-\frac { A }{ R } \frac { d }{ dt } ({ B }_{ 0 }sin\omega t)=-\frac { { AB }_{ 0 } }{ R } cos\omega t(\omega )\\ =-\frac { { A\omega B }_{ 0 } }{ R } cos\omega t\)

No, a stationary charge does not produce magnetic field.