\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)

λ = 5500 x 10^-10 m; P =150 W; Efficiency =12%
\({ E }=\cfrac { hc }{ { \lambda } } =\cfrac { 6.626\times { 10 }^{ -34 }\times 3 \times { 10 }^{ 8 } }{ 5500\times { 10 }^{ -10 } }=3.614 \times 10^{-19} J \)
\(n=\frac{E}{hv}=\cfrac { 150 }{ 3.614\times { 10 }^{ -19 } } =4.15 \times 10^{20}\times\frac{12}{100}\)
n = 4.98 x 10¹⁹s^-1

\(\mathrm{V}_{\mathrm{cc}}=12 \mathrm{~V}, \mathrm{R}_{\mathrm{B}}=10 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{E}}=1 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{c}}=1+1=2 \mathrm{k} \Omega, \alpha=0.99, \mathrm{~V}_{\mathrm{BE}}=0.7 \mathrm{~V}, \mathrm{I}_{\mathrm{c}}=?\)
\(\beta=\alpha /(1-\alpha)=0.99 /(1-0.99)=99\)
\(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}} / \beta=\mathrm{I}_{\mathrm{c}} / 99\)
Applying Kirchoff's Voltage law,
\(I_C R_C+I_n R_n+I_E R_E+V_{u t}=V\)
\(2 \times 10^3 \mathrm{I}_{\mathrm{C}}+10 \times 10^3\left(\mathrm{I}_{\mathrm{C}} / 99\right)+1 \times 10^3\left(\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{C}} / 99\right)+0.7=12 \quad\left(\because \mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{n}}+\mathrm{I}_{\mathrm{C}}\right)\)
\(\therefore \mathrm{I}_{\mathrm{C}}=\frac{11.3 \times 10^{-3} \times 99}{298}\)
\(\mathrm{I}_{\mathrm{C}}=3.7 \times 10^{-3} \mathrm{~A}=3.7 \mathrm{~mA}\)

Given data:
λ = 900 nm = 900 x 10^-9 m
λ ₂ = 750 nm = 750 x 10^-9 m
D = 2 m d = 2 nm = 2 x 10^-3 m
Let n^th order bright fringe of λ₁,
Coincides with (n + 1)^th order bright fringe of λ ₂
\(y_{n}=\frac{n \lambda_{1} D}{d}, Y_{n+1}=\frac{(n+1) \lambda_{2} D}{d} \)
\(\frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \)
\(n \lambda_{1}=(n+1) \lambda_{2} \)
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{900 \times 10^{-9}}{750 \times 10^{-9}}=\frac{18}{15}=\frac{6}{5} \)
\(1+\frac{1}{n} =\frac{6}{5} \)
\(\frac{1}{n} =\frac{6}{5}-1=\frac{6-5}{5} \)
\(\frac{1}{n} =\frac{1}{5} \)
n = 5
n + 1 = 6
5^th bright fringe of  λ ₁ coincides with 6^th bright fringe of λ ₂ in the least distance of y
\(y =\frac{n \lambda_{1} D}{d}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}} \)
\(=4500 \times 10^{-6}=4.5 \times 10^{-3} \)
y = 4.5 mm

r - Radius of surface of water
h - Depth of the water
\(\sin c=\frac{1}{n} \)
\(n=\frac{4}{3} \)
h = 80 cm = 0.80 m
\(\sin c =\frac{1}{\frac{4}{3}}=\frac{3}{4} n=\frac{1}{\sin c} \)
\(\tan c =\frac{\sin c}{\cos c} \)
\(=\frac{\sin c}{\sqrt{1-\sin ^{2} c}} \)
\(=\frac{1}{n \sqrt{1-\frac{1}{n^{2}}}} \)
\(\tan c =\frac{1}{\sqrt{n^{2}-1}} \)
\(\frac{r}{h} =\frac{1}{\sqrt{n^{2}-1}} ; r=\frac{h}{\sqrt{n^{2}-1}} \)
\(=\frac{0.80}{\sqrt{\left(\frac{4}{3}\right)^{2}}-1} \)
\(=\frac{0.80}{\sqrt{\frac{16}{9}-1}}=\frac{0.80 \times 3}{\sqrt{7}} \)
\(=\pi r^{2}=\frac{22}{7} \times \frac{0.80 \times 3}{\sqrt{7}} \times \frac{0.80 \times 3}{\sqrt{7}} \)
\(=\frac{22}{49} \times 2.4 \times 2.4 \)
\(=\frac{22 \times 5.76}{49}=\frac{126.72}{49}=2.586 \)
A = 2.6 m²

M = 100 g = 100 x 10^-3 kg
M = 0.1 kg
g = 10 m/s²
N = 5
B = 0.5 T
R = 20 cm = 20 x 10^-2 m
When, the sphere is in translational equilibrium
\(\mathrm{f}_{\mathrm{s}}-\mathrm{Mg} \sin \theta =0 \ldots(1) \)
\(\mathrm{f}_{\mathrm{s}} =\mathrm{Mg} \sin \theta \)
When, the sphere is in rotational equilibrium,
Torque ፒ = μB sin θ
For Equilibrium, f_sR - ፒ = 0
\(f_{s} R-\mu B \sin \theta=0 \ldots(2)\)
eqn (1) subs. eqn (2)

\(M g \sin \theta R-\mu B \sin \theta =0 \)
\(\sin \theta(M g R-\mu B) =0 \)
\(M g R =\mu B \)
\(\text { Now } \mu =N I \pi R^{2} \)
\(\therefore \operatorname{MgR} =N I \pi R^{2} B \)
\(\therefore I =\frac{M g}{N \pi R B} \)
\(I =\frac{0.1 \times 10}{5 \times \pi \times 20 \times 10^{-2}\times 0.5 }\)
\(=\frac{1}{\pi \times 50 \times 10^{-2}}=\frac{1 \times 10^{2}}{\pi \times 50}=\frac{100}{\pi \times 50}\)
\(\mathbf{I} =2 / \pi \mathrm{A} \)

Magnetic field \(\overset { \rightarrow }{ B } ={ 0.500\hat { i } T } \)
Velocity of the particle
\(\hat { v } \) = (1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) ms^-1
Charge of the proton q = 1.67 x 10^-19 C
Mass of the proton m = 1.67 x 10^-27kg
(a) The force experienced by the proton is \(\overset { \rightarrow }{ F } \) = q(\(\overset { \rightarrow }{ v } \) x \(\overset { \rightarrow }{ B } \) )
= 160 x 10^-19 x  ((1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) x (0.500 \(\hat { i } \)))
\(\overset { \rightarrow }{ F } \)= 1.60 x 10^-14 \(\hat { j } \)  N
Therefore, from Newton’s second law,
\(\overset { \rightarrow }{ a } =\frac { 1 }{ m } \overset { \rightarrow }{ F } =\frac { 1 }{ 1.67\times { 10 }^{ -27 } } (1.60\times { 10 }^{ -14 })\hat j\)
\(=9.58\times { 10 }^{ 12 }\hat jm{ s }^{ -2 }\)
(b) Trajectory is helical Radius of helical path is
\(R=\frac { { mv }_{ z } }{ \left| q \right| B } =\frac { 1.67\times { 10 }^{ -27 }\times 2.00\times { 10 }^{ 5 } }{ 1.60\times { 10 }^{ -19 }\times 0.500 } \)
= 4.175 x 10^-3m = 4.18mm
Pitch of the helix is the distance travelled along x-axis in a time T, which is P = v_x T
But time, \(T=\frac { 2\pi }{ \omega } =\frac { 2\pi m }{ \left| q \right| B } =\frac { 2\times 3.14\times 1.67\times { 10 }^{ -27 } }{ 1.60\times 1{ 0 }^{ -19 }\times 0.500 } =13.1\times { 10 }^{ -8 }s\)
Hence, pitch of the helix is
\(p={ v }_{ x }T=(1.95\times { 10 }^{ 5 })(13.1\times { 10 }^{ -8 })=25.5\times { 10 }^{ -3 }m=25.5mm\)
The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.

(i) Electric field is given by \(|E|=\frac{d V}{d x}\)
(a) For region A,
dV = 8 - 5 = 3V, dx - (0.2 - 0) = 0.2m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{3}{0.2}=15 \mathrm{Vm}^{-1}\)
(b) For region B,
dv = 0, E_x = 0
(c) For region C
dV = (7 - 5) = 2V, dx = (0.6 - 0.4) = 0.2 m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{2}{0.2}=10 \mathrm{Vm}^{-1}\)
(d) For region D
dV = (7 - 1)=6V,  dx = (0.8 - 0.6) = 0.2 m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{6}{0.2}=30 \mathrm{Vm}^{-1}\)
(ii) Electric field as a function of x for Figure (ii).

From the figure F₁ = F₅ but opposite in direction. So cancel each other
Also, F₂ sin 45^o = F₄ sin 45^o but opposite in direction. So cancel each other.

\( \mathrm{F}_{\mathrm{tot}} =\mathrm{F}_{3}+\mathrm{F}_{2} \cos 45^{\circ}+\mathrm{F}_{4} \cos 45^{\circ} \)
\(\mathrm{F}_{\mathrm{tot}} =\frac{k q Q}{R^{2}}+\frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}}+\frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}} \)
\(\mathrm{~F}_{\mathrm{tot}} =\frac{k q Q}{R^{2}}+2 \frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}} \)
\(=\frac{k q Q}{R^{2}}+\sqrt{2} \sqrt{2} \frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}} \)
\(\vec{F}_{\text {tot }} =\frac{k q Q}{R^{2}}[1+\sqrt{2}] \hat{i} N\)
\(\vec{F} =\frac{1}{4 \pi \varepsilon_{o}} \frac{q Q}{R^{2}}(1+\sqrt{2}) \hat{i N}\)

(a) The capacitance of the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } =\frac { 8.85\times { 10 }^{ -12 }\times 25\times 10^{ -4 } }{ 1\times 10^{ -3 } } \)
= 221.2 x 10^-13 F
C = 22.12 x 10^-12F = 22.12pF
(b) The charge stored in any one of the plates is Q = CV, Then
Q = 22.12 x 10^-12 x 10 = 221.2 x 10^-12C = 221.2 pC.

(a) The power delivered by the battery P = VI. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current (P ∝ I). So 60 W bulb draws twice as much as current as 30 W and it will glow brighter than 30 W bulb.
(b) To calculate the resistance of the bulbs, we use the relation \(P=\frac { { v }^{ 2 } }{ R } \) In both the bulbs, the voltage drop is the same, so the power is inversely proportional to the resistance or resistance is inversely proportional to the power \(\left( R∝ \frac { 1 }{ P } \right) \). It implies that, the 30W has twice as much as resistance as 60 W bulb.
(c) When these two bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it depends whether bulbs are connected in series or parallel.

P = 50 mW; λ = 640nm = 640 x 10^-9 m
P = 50 x 10^-3W
\(n=\cfrac { hc }{ \lambda } = \frac{6.626 \times10^{-34} \times 3 \times 10^8}{640 \times 10{-9}}=3.106 \times 10^{-19}J\)
\(n=\frac{E}{hv}=\cfrac { 50\times { 10 }^{ -3 } }{ 3.106\times { 10 }^{ -19 } } = 1.61\times 10^{17} s^{-1}\) 
n = 1.61 x 10¹⁷ s^-I

\(T_{1 / 2}=1600 \text { years }=1600 \times 365 \times 24 \times 60 \times 60 s\)
R = 1 curie = 3.7 x 10¹⁰ Bq, Show that m = 1g
R = λN
Number of atoms Present, N = \(\frac{\mathrm{R}}{\lambda}=\frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2}\)
Mass of 6.023 x 10²³ atoms of  \({ }_{88}^{226} R a=226 g\)
Mass of 1 atom of  \({ }_{88}^{226} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \mathrm{~g}\)
Mass of N atoms of \({ }_{88}^{{ }{266}} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \times \mathrm{Ng}\)
Mass of N atoms of \({ }_{88}^{226} \mathrm{Ra}(\mathrm{m})=\frac{226}{6.023 \times 10^{23}} \times \frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2} \mathrm{~g}\)
\(\mathrm{m}=\frac{226}{6.023 \times 10^{23}} \times \frac{3.7 \times 10^{10}}{0.6931} \times 1600 \times 365 \times 24 \times 60 \times 60 \mathrm{~g}\)
m = 1.01 g

Magnetic field B = 0.2 T
Current flowing through the coil I = 3A
Cross sectional area of a circular coil A =  0.1 cm²
∴ A = 0.1 x 10^-4 m²
(a) Total torque on the coil:
\(\tau=\text { IBA } \cos \theta\)
Here, the plane of the loop is perpendicular to the direction of magnetic field
\(\therefore \theta=90^{\circ} \)
\(\therefore \tau=\text { IBA } \cos 90^{\circ} \)    (∵ cos90^o = 0)
= 3 x 0.2 x 0.1 x 10^-4 x 0
= zero
∴ Total Torque on the coil = zero
(b) Total force on the coil \(F=B q v \sin \theta \)
\(Here\ \theta=0 \)
\(\therefore F=0.2 \times \mathrm{qv} \times \sin \theta=\text { zero } \)
Total force on the coil = zero
c) Average force: F = Bqv
Charge density \(=\sigma=\frac{q}{A} \)
∴ Charge  \(q =\sigma A \)
\(=10^{28} \times 0.1 \times 10^{-4} \)
\(q=10^{23} \mathrm{C} \)
Force = Bll and I = 1 m
F = 0.2 x 3 x 1 = 0.6 N
∴ Average force on each electron
\(=\frac{F}{q}=\frac{0.6}{10^{23}} \)
\(F =0.6 \times 10^{-23} \mathrm{~N} \)

Magnitude of induced \(\mathrm{emf}|\mathrm{e}|=\left|-\frac{d \phi}{d t}\right|\)
\(\therefore c=\frac{d \phi}{d t}\)
(i) In the region ab,
\(c_1=\frac{d \phi_1}{d t}=\frac{4-0}{1-0}=\frac{4}{1}=4 \mathrm{~V}\)
(ii) In the region bc,
\(c_2=\frac{d \phi_2}{d t}=\frac{4-4}{3-1}=\frac{0}{2}=0 \mathrm{~V}\)
(iii) In the region cd,
\(e_3=\frac{d \phi_2}{d t}=\frac{4-2}{4-3}=\frac{2}{1}=2 \mathrm{~V}\)
(iv) In the region de,
Hence,
\(e_4=\frac{d \phi_4}{d t}=\frac{2-0}{7-4}=\frac{2}{3}=0.66 V  \)
\(\mathrm{e}_2<\mathrm{e}_4<\mathrm{e}_3<\mathrm{e}_1\)
In the ascending order of the magnitude of induced emf
Region bc < Region de < Region cd < Region ab.

Current, I = 4 A, Magnetic flux, \(\phi=6 \times 10^{-5} \mathrm{~Wb}\), Number of turns, N = 200
Self inductance, \(L=\frac{N \phi}{I}\)
\(\therefore L  =\frac{200 \times 6 \times 10^{-5}}{4}  \)
\(L  =300 \times 10^{-5} \mathrm{H}=3 \times 10^{-3} \mathrm{H}\)
∴ Magnetic energy, \(U_s=\frac{1}{2} L I^2\)
\(U_s  =\frac{1}{2} \times 3 \times 10^{-3} \times(4)^2  \)
\(=\frac{1}{2} \times 3 \times 10^{-3} \times 16  \)
\(=24 \times 10^{-3} \mathrm{~J}=0.024 \mathrm{~J}\)
Magnetic energy = 0.024 J

Velocity of the particle is \(\vec { v } =v\hat { j } \)
(a) Magnetic field is along positive y-direction, this implies \(\vec B=B\hat { j } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { j } )=\vec 0\)
So, no force acts on the particle when it moves along the direction of magnetic field.
(b) Since the magnetic field points in positive z - direction, this implies, \(\vec { B } =B\hat { k } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { k } )=qvB\vec i \)
Therefore, the magnitude of the Lorentz force is qvB and direction is along positive x - direction.
(c) Magnetic field is in zy - plane and making an angle θ with the velocity of the particle, which implies \( {\vec B } =Bcos\theta \hat { j } +Bsin\theta \hat { k } \)

From Lorentz force,
\({ \vec F }_{ m }=q(v\hat { j } )\times (Bcos\theta \hat { j } +Bsin\theta \hat k)\) 
\(=qvBsin\theta \hat { i } \)

Resistivity of the first material \({ \rho }_{ 1 }=4\times { 10 }^{ -3 }\Omega m\)
Length of the first material, l₁ = 25 x 10² m   
Area of cross section of the first material, A₁ = 3 x 3 x 10^-6 
= 9 x 10^-6 m²
Resistivity of second material, \({ \rho }_{ 2}=5\times { 10 }^{ -3 }\Omega m\) 
Length of second material, \(l_{ 2 }=70\times { 10 }^{ -2 } m\) 
Area of cross section second material, A₂ = 3 x 3 x 10^-6
= 9 x 10^-6 m²
Resistivity of first material, \(\rho_{1}= R_{1} \times \frac {A_{1}}{l_{1}}\)  
Resistance first material, \(R_{1} = \frac {\rho_{1}l_1}{A_{1}}\) 
\(\therefore R_1=\frac{4 \times 10^{-3} \times 25\times 10^{-2}}{9 \times10^{-6}}\)
\(R_1=\frac{100}{9}\times10^{-5+6}=\frac{1000}{9}\Omega\)
Resistivity of second material, \(\rho_{2}=\frac{R_{2} A_{2}}{l_{2}} \)
∴ Resistance of second material, \(R_{2}=\frac{\rho_{2} l_{2}}{A_{2}} \)
\(\therefore R_{2} =\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}} \)
\(R_2=\frac{350 \times 10^{-5}}{9 \times 10^{-6}}=\frac{350}{9} \times 10 \)
\(R_2=\frac{3500}{9} \Omega \)
Total resistance \(R_{t} =\frac{1000}{9}+\frac{3500}{9} \)
\(=\frac{4500}{9}=500 \Omega \)
\(\therefore\) Total resistance = 500 \(\Omega\)

Area of cross section  A = 10^-6 m² Current I = 2A
Number of electrons per cubic metre, n = 8 \(\times\)10²⁸
Current I = nAev_d
∴ Drift Velocity \(v_{d}=\frac{I}{n A e}\)
\(v_{d} =\frac{2}{8 \times 10^{28} \times 10^{-6} \times 1.6 \times 10^{-19}} \)
\(=\frac{2}{8 \times 10^{3} \times 1.6} \)
\(=\frac{20}{128} \times 10^{-3}=15.6 \times 10^{-5} \mathrm{~ms}^{-1} \)
Current density J = l / A.
\(J=\frac{2}{10^{-6}}\)
J = 2 x 10⁶ Am^-2

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.

\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 2 }{ f } \right) \) 
\(\cfrac { 1 }{ 2 } =\left( \mu -1 \right) \) 
\(\mu =1.5\)

(i) The satellite communication is a mode of transmission of signal between transmitter and receiver via satellite.
(ii) The message signal from the Earth station is transmitted to the satellite on board via an uplink (frequency band 6 GHz), amplified by a transponder and then retransmitted to another earth station via a downlink (frequency band 4 GHz).
Applications:
Satellites are classified into different types based on their applications.
(i) Weather Satellites:
They are used to monitor the weather and climate of Earth. By measuring cloud mass, these satellites enable us to predict rain and dangerous storms like hurricanes, cyclones etc.
(ii) Communication satellites:
They are used to transmit television, radio, internet signals etc. Multiple satellites are used for long distance communication
(iii) Navigation satellites:
These are employed to determine the geographic location of ships, aircraft or any other object.

Magnetic force on the wire,
F BIl sinθ
= BIl sin90°       (∵ θ = 90⁰)
= 0.27 x 10 x 3 x 10^-2
F = 8.1 x 10^-2 N

Given:
Induced emf, e = Blv (or) e = \(\frac12\) Bl²ω [where, v = rω, 1 = 2R]
Also, ω=2ㅠf = 2π x \(\frac{120}{60}\)
ω = 4ㅠ
Solution:
emf, e = \(\frac12\) x 4ㅠ x 0.4 x 10^-4 (0.5)²
= 6.28 x 10^-5V
The number of spokes is immaterial because the emf's across the spokes are in parallel