N = 5 turns
A = 2 x 10^-2 m²
B = 4 x 10^-2 Wb m^-2
K = 4 x 10^-9 N m deg^-1
(a) Current sensitivity
\({ I }_{ s }=\frac { NAB }{ K } =\frac { 5\times 2\times { 10 }^{ -2 }\times 4\times { 10 }^{ -2 } }{ 4\times 10^{ -9 } } \)
= 10⁶ divisions per ampere
\(I\mu A=\) 1microambire =10^-6ampere
Therefore,
\({ I }_{ s }={ 10 }^{ 6 }\frac { div }{ A } =1\frac { div }{ { 10 }^{ -6 }A } =1\frac { div }{ \mu A } \)
\({ I }_{ s }=1div{ \left( \mu A \right) }^{ -1 }\)
(b) Voltage sensitivity
\({ V }_{ s }=\frac { \theta }{ V } =\frac { 50div }{ 25mv } =2\times { 10 }^{ 3 }{ div \ V }^{ -1 }\)
(c) The resistance of the galvanometer is
\({ R }_{ g }=\frac { { I }_{ s } }{ { v }_{ s } } =\frac { { 10 }^{ 6 }\frac { div }{ A } }{ { 2\times }10^{ 6 }\frac { div }{ V } } =0.5\times { 10 }^{ 3 }\frac { V }{ A } =0.5k\Omega \)

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 = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

R_s = R₁ + R₂ = 15 Ω           (1)
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }{ +R }_{ 2 } } =\frac { 56 }{ 15 } \Omega \quad \) (2)
From equation (1) substituting for R₁ + R₂ in equation (2)
\(\frac { { R }_{ 1 }{ R }_{ 2 } }{ 15 } =\frac { 56 }{ 15 } \Omega \)
∴ R₁R₂ = 56
\({ R }_{ 2 }=\frac { 56 }{ 15 } \Omega \)         (3)
Substituting for R₂ in equation (1) from equation (3)
\({ R }_{ 1 }+\frac { 56 }{ { R }_{ 1 } } =15\)
Then, \(\frac { { R }_{ 1 }^{ 2 }+56 }{ { R }_{ 1 } } =15\)
R₁² + 56 = 15 R₁
R₁² - 15 R₁ + 56 = 0
The above equation can be solved using factorisation.
R₁ = 8 Ω (or) R₁ = 7 Ω
If (R₁ = 8 Ω)
Substituting in equation (1)
8 + R₂ = 15
R₂ = 15 – 8 = 7 Ω ,
R₂ = 7 Ω i.e , (when R₁ = 8 Ω ; R₂ = 7 Ω)
If R₁= 7 Ω
Substituting in equation (1)
7 + R₂ = 15
R₂ = 8 Ω , i.e , (when R₁ = 7 Ω ; R₂ = 8 Ω )

(a) Electric potential at point P is given by
V_p=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ p } } =\frac { 9\times 10^{ 9 }\times 9\times { 10 }^{ -6 } }{ 10 } \)  = 8.1 x 10³ V
Electric potential at point Q is given by
V_Q = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ Q } } =\frac { 9\times 10^{ 9 }\times 9\times { 10 }^{ -6 } }{ 16 } \) = 5.06 x 10³ V
Note that the electric potential at point Q is less than the electric potential at point P. If we put a positive charge at P, it moves from P to Q. However if we place a negative charge at P it will move towards the charge +9μC.
The potential difference between the points P and Q is given by
ΔV = V_p-V_Q = +3.04 x 10³ V
b) Suppose we replace the charge +9 μC by -9 μC, then the corresponding potentials at the points P and Q are,
V_p = -8.1 x 10³ V, V_Q = -5.06 x 10³ V
Note that in this case electric potential at the point Q is higher than at point P.
The potential difference or voltage between the points P and Q is given by
ΔV = V_p-V_Q= -3.04 x 10³V
(c) The electric potential V at a point P due to some charge is defined as the work done by an external force to bring a unit positive charge from infinity to P. So to bring the q amount of charge from infinity to the point P, work done is given as follows.
W = qV
W_Q = 2 x 10^-6 x 5.06 x 10³J = 10.12 x 10^-3 J.

In the first case, the resistance of the block
\({ R }_{ AB }=\rho \frac { length }{ Area } =\rho \frac { C }{ AB } \)
The current \({ I }_{ AB }=\frac { V }{ { R }_{ AB } } =\frac { V }{ \rho } .\frac { AB }{ C } \quad (1)\)
In the second case, the resistance of the block \({ R }_{ BC }=\rho \frac { A }{ BC } \)
The current \({ I }_{ BC }=\frac { V }{ { R }_{ BC } } =\frac { V }{ \rho } .\frac { BC }{ C } \quad (2)\)
To express I_BC interms of I_AB, we multiply and divide equation (2) by AC, we get
\({ I }_{ BC }=\frac { V }{ \rho } .\frac { BC }{ A } \frac { AC }{ AC } =\left( \frac { V }{ \rho } .\frac { AB }{ C } \right) .\frac { { C }^{ 2 } }{ { A }^{ 2 } } =\frac { { C }^{ 2 } }{ { A }^{ 2 } }.{ I }_{ AB }\)
Since C > A, the current I_BC > I_AB

Case (a)
The magnitude of the electric field at point P is
E_p = \(\frac { 1 }{ 4\pi \varepsilon _{ 0 } } \frac { q }{ { r }^{ 2 } } =\frac { 9\times { 10 }^{ 9 }\times 1\times 10^{ -6 } }{ 4 } \)
= 2.25 x 10³ NC^-1
Since the source charge is positive, the electric field points away from the charge. So the electric field at the point P is given by
\(\bar { { E }_{ p } } \) = 2.25 x 10³ NC^-1
For the point Q
\(|\vec { { E }_{ Q } } |=\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ 16 } \) = 0.56 x 10³ NC^-1
Hence \(\vec { { E }_{ Q } } \) = 0.56 x 10³\(\hat { j } \) NC^-1
Case (b)
The magnitude of the electric field at point P
\(\bar { { E }_{ p } } =\frac { kq }{ r^{ 2 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 2\times 10^{ -6 } }{ 4 } \)
= 4.5 x 10³ NC^-1
Since the source charge is negative, the electric field points towards the charge. So the electric field at the point P is given by
\(\vec { E_{ p } } \) = -4.5 x 10³\(\hat { i } \)NC^-1
For the point Q, \(|\vec { { E }_{ Q } } |\frac { 9\times 10^{ 9 }\times 2\times { 10 }^{ -6 } }{ 36 } \)
= 0.5 x 10³ NC^-1
\(\vec { E_{ Q } } \) = 0.5 x 10³\(\hat { i } \)NC^-1
At the point Q the electric field is directed along the positive x-axis.

If the two spheres are neutral, the angle between them will be 0^o when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical. At equilibrium, each charge experiences zero net force in each direction. We can draw a free-body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions.
The free-body diagram is shown below

In the x-direction, the acceleration of the charged sphere is zero.
Using Newton’s second law \((\vec { { F }_{ tot }= } m\vec { a } )\), we have
T sinθ\(\hat { i } \) - F_e\(\hat { i } \) =0
T sinθ = F_e    ......(1)
Here T is the tension acting on the charge due to the string and F_e is the electrostatic force between the two charges.
In the y-direction also, the net acceleration experienced by the charge is zero
Tcosθ\(\hat { j } \) - mg\(\hat { j } \) = 0
Tcosθ = mg   ..(2)
By dividing equation (1) by equation (2),
tanθ = \(\frac { { F }_{ e } }{ mg } \)          .....(3)
Since they are equally charged, the magnitude of the electrostatic force is
\({ F }_{ e }=k\frac { { q }^{ 2 } }{ { r }^{ 2 } } \) where k=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \)
Here r = 2a = 2Lsinθ. By substituting these values in equation (3),
tanθ = k\(\frac { { q }^{ 2 } }{ mg(2Lsin\theta )^{ 2 } } \)       ..........(4)
Rearranging the equation (4) to get q
q = 2 Lsinθ\(\\ \sqrt { \frac { mgtan\theta }{ k } } \)
= 2 x 0.1 x sin 30^o x \(\sqrt { \frac { 10^{ -3 }\times 10\times { tan30 }^{ 0 } }{ 9\times 10^{ 9 } } } \)
q = 8.01 x 10^-8C = 80.1 nC

\(\beta =\cfrac { D\lambda }{ d } ,i.e\beta \propto \lambda \)  the wavelength of blue light is less than that of red light, hence if red light is replaced by blue light, the fringe width decreases, i.e., fringes come closer.

Industrial robots are used for welding, cutting, robotic water jet cutting, robotic laser cutting, lifting, sorting, bending, manufacturing, assembling, packing, transport, handling hazardous materials like nuclear waste, weaponry, laboratory research, mass production of consumer and industrial goods.

A P-N junction diode which converts an optical signal into electric current is known as photo diode.

The fuel bundles which consist of tiny pellets of uranium oxide are placed in calandria reactor vessel. The part of the reactor vessel which contains the fuel rod is known as reactor core.

In nuclear fusion reaction, the energy liberated per unit mass of the nuclei taking part in the reaction is many times larger than the energy liberated in a fission reaction.

A nuclear reaction in which the neutron used to carry out the nuclear fission reaction gets multiplied as more and more such fission reaction take place is called a nuclear chain reaction.

Linear mass density of the conductor is = 0.2 g/m
Mass per unit length \(\frac{M}{l}=0.2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\)
Magnetic field B = 1T.
Acceleration due to gravity, g = 10 ms^-2
Force \(=\frac{m}{l} \times g\)
= 0.2 x 10^-3 x 10 = 0.2 x 10^-2
F = 2 x 10^-3 N   ....(1)
If the coil is placed in the magnetic field then the force acting on the coil is
F= BIl   ....(2)
From the equation (1) and (2) we get
BIl = 2 x 10^-3
∴ 1 x L x I = 2 x 10^-3
∴ I = 2 x 10^-3 A      [∴ l = 1m]
∴ I = 2mA

Electric field, E = 6.0 x 10⁶ N C^-1 and magnetic field, B = 0.83 T.
Then.
\(v=\frac { E }{ B } =\frac { { 6.0\times 10 }^{ 6 } }{ 0.83 } =7.23\times { 10 }^{ 6 }{ ms }^{ -1 }\)
When an electron goes with this velocity, it shows null deflection. Since the accelerating potential is 200 V, the electron acquires kinetic energy because of this accelerating potential. Hence,
\(\frac { 1 }{ 2 } mv^{ 2 }=eV \)
\(v=\sqrt { \frac { 2eV }{ m } }\)
Since the mass of the electron, m = 9.1 x 10⁻³¹kg and charge of an electron, \(\left| q \right| =e=1.6\times { 10 }^{ -19 }C.\) The velocity acquired by the electron due to accelerating potential 200 V is
\({ v }_{ 200 }=\sqrt { \frac { 2\left( 1.6\times { 10 }^{ -19 } \right) \left( 200 \right) }{ \left( 9.1\times { 10 }^{ -31 } \right) } } =8.39\times { 10 }^{ 6 }m{ s }^{ -1 }\)
Since the speed v₂₀₀ > v, the electron is deflected towards direction of Lorentz force. So, in order to have null deflection, the potential, we have to supply is
\(v=\frac { { 1mv }^{ 2 } }{ 2\quad e } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 7.23\times { 10 }^{ 6 } \right) ^{ 2 } }{ 2\times \left( 1.6\times { 10 }^{ -19 } \right) } \)
V = 148.65 V

(a) \(C_{1}=\frac{\varepsilon_{0} A}{2d} (\because area=\frac{A}{2})\)
\(C_{2}=\frac{\varepsilon_{r} \varepsilon_{0} A}{d} \)
C₁ and C₂ are in parellel,
\(C_{P}=\left(C_{1}+C_{2}\right) \)
\(C_{P}=\frac{1}{2}\left(\frac{\varepsilon_{r} \varepsilon_{0} A}{d}+\frac{\varepsilon_{0} A}{d}\right) \)
\(=\frac{1}{2} \frac{\varepsilon_{0} A}{d}\left(\varepsilon_{r}+1\right) \)
\(C_{P}=\frac{\varepsilon_{0} A}{2 d}\left(\varepsilon_{r}+1\right) \)

(b) \(C_1=\frac{\varepsilon_0 \varepsilon_r A}{d / 2}=\frac{2 \varepsilon_0 \varepsilon_r A}{d}(\because \text { Separation }=d / 2)\) 
\(C_2=\frac{2 \varepsilon_0 A}{d}\)
C₁ and C₂ are in series,
\(\frac{1}{C_1}  =\frac{1}{C_1}+\frac{1}{C_2}  \)
\(=\frac{d}{2 \varepsilon_0 \varepsilon_r A}+\frac{d}{2 \varepsilon_0 A}  \)
\(=\frac{d}{2 \varepsilon_0 A}\left(\frac{1}{\varepsilon_r}+1\right)  \)
\(\frac{1}{C_1}  =\frac{d}{2 \varepsilon_0 A}\left(\frac{1+\varepsilon_r}{\varepsilon_r}\right)  \)
\(\therefore C_s  =\frac{2 e_0 A}{d}\left(\frac{e_r}{1+e_r}\right)\)

\(W =\left(V_{b}-V_{a}\right) q\left[where\ V=\frac{K Q}{r}\right] \)
To find V_b :
\(V_b=\frac{kQ}{r_3}+\frac{kQ}{r_4}\)
\(V_{b} =\frac{K \times 10 \times 10^{-6}}{\sqrt{50 \times 10^{-4}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{250 \times 10^{-4}}} \)
\(V_{b} =\frac{K \times 10^{-5}}{\sqrt{50} \times 10^{-2}}+\frac{K \times 10^{-5}}{\sqrt{250} \times 10^{-2}} \)
\(=K \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
\(=9 \times 10^{9} \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
V_b = 1842002 V
To find V_a :
\(V_a=\frac{kQ}{r_1}+\frac{kQ}{r_2}\)
\(V_{a} =\frac{K \times 10 \times 10^{-6}}{\sqrt{5 \times 10^{-2}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{15 \times 10^{-2}}} \)
V_a = 2400000 V
\(\therefore W_{D} =\left(V_{b}-V_{a}\right) q=(1842002-2400000)( -2 \times 10^{-6} )\)
W = +1.12J
Positive sign implies that to move the charge - 2 μC external work is required.

(a) In the figure

1 - Electrostatic force F_e = QE
2 - Weight W = mg
3 - Elastic force F = -kx
4 - Upward force = Normal reaction = N
b) In this figure

1- Electrostatic force F = qE
2 - Weight W = mg
3 - Tension acting along the string is T
(c) The charge is attracted towards the positively charged plate because it is a negative charge.

1 - Force = qE
2 - Downward force F = mg

Note: A similar problem is solved in XIth Physics volume I, unit 3 section 3.3.2. There are three forces that acts on the mass m:
(i) The downward gravitational force exerted by the Earth (mg)
(ii) The normal force exerted by the inclined surface (N)
(iii) The Coulomb force given by uniform electric field (qE) The free body diagram for the mass m is drawn below.

A convenient inertial coordinate system is located in the inclined surface as shown in the figure. The mass m has zero net acceleration both in x and y-direction.
Along x-direction, applying Newton’s second law, we have
mg sinθ\(\hat { i } \) - qE\(\hat { i } \) = 0
mg sinθ - q E = 0
or, E = \(\\ \frac { mgsin\theta }{ q } \)
Note that the magnitude of the electric field is directly proportional to the mass m and inversely proportional to the charge q. It implies that, if the mass is increased by keeping the charge constant, then a strong electric field is required to stop the object from sliding. If the charge is increased by keeping the mass constant, then a weak electric field is sufficient to stop the mass from sliding down the plane.
The electric field also can be expressed in terms of height and the length of the inclined surface of the plane.
E = \(\frac { mgh }{ qL } \).

Let the currents flowing through wires XY and PQ be I₁ and I₂
Magnetic field along PQ is \(\mathrm{B}_{1}=\frac{\mu_{o} I_{2}}{2 \pi r}\)
Force per unit length on PQ is \(\frac{F_{2}}{l}=\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)
If the wire XY is slightly displaced and released, it executes simple harmonic motion with the condition that acceleration is directly proportional to the displacement y
\(\therefore a=-\omega^{2} y\)  .....(1) 
The distance between two wires = d
Time period
\(T=\frac{2 \pi}{\omega} \)
\(a=\frac{g}{d} y \) .....(2)
By comparing the equations (1) and (2) we get
\(\omega^{2} =-\frac{g}{d} \ \therefore \omega=\sqrt{\frac{g}{d}} \)
\(\text { Time period } =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{d}{g}} \)
\(\therefore T =2 \pi \sqrt{\frac{d}{g}} \)

According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.
Consider the point at which the total potential zero is located at a distance x from the charge +q as shown in the figure.

The total electric potential at P is zero.
V_tot = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ x } -\frac { 2q }{ (9-x) } \right) \)=0
Which gives \(\frac { q }{ x } -\frac { 2q }{ (9-x) } \)
or \(\frac { 1 }{ x } =\frac { 2 }{ (9-x) } \)
Hence, x = 3m

Magnetic moment,
μ = IA
Angular momentum,
L = Iω
Ratio \(\frac{p_m}{L}=\frac{(q/T)\pi r^2}{mr^2\omega}=\frac{q}{2m}\)

Lorentz force F = Bqv
Energy w = qV
Energy is equal to kinetic energy,
\(qV=\frac{1}{2}mv^2\)
\(v=\sqrt { \frac {2qV }{ m } } \)
\(\therefore Lorentz \ force \ F= Bq\times \sqrt \frac{2qV}{m}=\sqrt { \frac { 2{ B }^{ 2 }{ q }^{ 3 }V }{ m } } \)

L = 20 x 10^-3H. C = 50 x 10^-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340  \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340  \)
\(=6800 \times 10^{-1}=6.8 \Omega  \)
\(X_C=\frac{1}{C .}  \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega   \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2}  \)
\(=\sqrt{(40)^2+(58.82-6.8)^2}  \)
\(=\sqrt{(40)^2+(52.02)^2}  \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62}  \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi  \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W

\(\omega=250 \mathrm{rad} / \mathrm{s}, C=4 \times 10^{-} \mathrm{F}  \)
\(R=100 \Omega, \quad \mathrm{V}_{\mathrm{R}}=40 \mathrm{~V}  \)
\(\therefore I_{\mathrm{R}}=\frac{V_R}{100}=\frac{40}{100}=0.4 \mathrm{~A}  \)
\(\omega=\frac{1}{\sqrt{L C}}  \)
\(\omega^2=\frac{1}{L C}  \)
\((250)^2=\frac{1}{L \times 4 \times 10^{-6}}  \)
\(L=\frac{1}{4 \times(250)^2 \times 10^{-6}}  \)
\(=\frac{1}{4 \times 250 \times 250 \times 10^{-6}}  \)
\(=\frac{1}{1000 \times 10^{-6} \times 250}  \)
\(=\frac{10^3}{250}=\frac{1000}{250}=4 \mathrm{H}\)
Voltage acnoss L, V_t = IX_L
V_L = l x L x ω
= 0.4 x 4 x 250
0.4 x 1000 = 400 V

In RL circuit, tanΦ = \(\frac{X_l}{R}\)
If R = X₁, then tanΦ = \(\frac{X_l}{X_l}=1\)
∴ Φ = tan^-1(1)=\(\frac{\pi}{4}\)
∴ Phase difference \(=\frac{\pi}{4}\)

\(\vec {E}\) = 10x\(\hat{i},\)  when x = 2 m
\(\vec {E}\) = 10 x 2 x \(\hat{i}\) = 20\(\hat{i}\)
Since, \(E=\frac{-dV}{dx}\therefore V=+20 V\)

\(U=\frac{1}{4\piε_0}\frac{q_1q_2}{r_{12}}\)
\(i) U=\frac{1}{4\piε_0}\frac{Q(-Q)}{r}=\frac{1}{4\piε_0}[\frac{-Q^2}{r}]\)
\(ii) U=\frac{1}{4\piε_0}\frac{(-Q)(-Q)}{r}=\frac{1}{4\piε_0}[\frac{Q^2}{r}]\)
\(iii) U=\frac{1}{4\piε_0}\frac{Q(2Q)}{r}=\frac{1}{4\piε_0}[\frac{2Q^2}{r}]\)
\(iv) U=\frac{1}{4\piε_0}\frac{Q(-2Q)}{2r}=\frac{1}{4\piε_0}[\frac{-Q^2}{r}]\)
From the values, 1 = 4 < 2 < 3

The charge + q will be stable between B₁ and B₂ with respect to the displacement.

I = 0.2 A, R = 10 Ω, E = 2.1 V
\(I=\frac{ɛ}{R+r}\)
\(0.2=\frac{2.1}{10+r}\)
0.2 x (10 + r) = 2.1
2 + 0.2 r = 2.1
0.2 r = 2.1 - 2 = 0.1
Internal resistance, \(r=\frac{0.1}{0.2}=\frac{1}{2}\)
r = 0.5 Ω

Brown - 1
Black - 0
Yellow - 10⁴
(∴ R = 10 x 10⁴ Ω = 100 kΩ)

Yellow - 4
Violet - 7
Orange - 10³
Silver - Tolerance - 10%

Resistance, \(R=\frac{V}{I}=\frac{4}{2}=2 \ ohm\)