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

For a charged particle moving on a magnetic field, \(\vec { F } \)= q(\(\vec { v } \)x\(\vec { B } \))
The work done by the magnetic field is
\(W=\int { \overset { \rightarrow }{ F } .{ d \vec r } =\int { \overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } dt } } \)
\(W=q\int { \left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) } .\overset { \rightarrow }{ v } dt=0\)
Since \(\overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \) is perpendicular to \(\vec { v } \) and hence \(\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) .\overset { \rightarrow }{ v } =\overset { \rightarrow }{ 0 } \) 
This means that Lorentz force does no work on the particle. From work-kinetic energy theorem, (Refer section 4.2.6, XI th standard Volume I)
\(\frac { dw }{ dt } =p=0\)
Since \(\overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } =0\Rightarrow \overset { \rightarrow }{ F } \ and \ \overset { \rightarrow }{ v } \) are perpendicular to each other. The angle between Lorentz force and velocity of the charged particle is 90^o. Thus Lorentz force changes the direction of the velocity but not the magnitude of the velocity. Hence Lorentz force does no work and also does not alter kinetic energy of the particle. 

The magnitude of deflecting torque (the torque which makes the object rotate) acting on the bar magnet will tend to align the bar magnet parallel to the direction of the uniform magnetic field \(\overset { \rightarrow }{ B } \) 
\(\left| \overset { \rightarrow }{ r } \right| ={ p }_{ m }Bsin\theta \)
The magnitude of restoring torque acting on the bar magnet can be written as
\(\left| \overset { \rightarrow }{ r } \right| =I\frac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } \)
Under equilibrium conditions, both magnitudes of deflecting torque and restoring torque will be equal but act in the opposite directions, which means
 \(\frac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } =-{ p }_{ m }Bsin\theta \)

Since the resistors are connected in series, the effective resistance in the circuit
= 4 Ω + 6 Ω = 10 Ω
The Current I in the circuit =\(\frac { V }{ { R }_{ eq } } =\frac { 24 }{ 10 } =2.4A\)
Voltage across 4Ω resistor
V₁= IR₁ = 2.4A x 4Ω = 9.6V
Voltage across 6 Ω resistor
V₂ = IR₂ = 2.4A x 6Ω = 14.4V

For getting a sharp image, we have to use light of suitable wavelength. This is because focal length of a lens depends upon wavelength.

Power of a lens increases if red light is replaced by violet light because \(\\ \\ P=\cfrac { 1 }{ f } \left( _{ a }{ n }_{ g }-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) and refractive index is maximum for violet light in visible region of spectrum.

The angular resolution has a unit in radian (rad) and it is given by the equation,
\(\theta =\cfrac { 1.22\lambda }{ \alpha } \)

The stopping potential does not depend on the intensity of incident radiation; so stopping potential will remain unchanged.

The scales on the wings of a morpho butterfly contain nanostructures that change the way light waves interact with each other, giving the wings brilliant metallic blue and green hues.

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.

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

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

B = 4.1 x 10^–5 T; θ = 90^o – 47^o = 43° ;
A = 3m²
We know that   \(\Phi_{B}=B A \cos \theta\) 
\(\Phi_{\mathrm{B}}\) = 4.1 x 10^–5 x 3 x cos 43^o  
= 4.1 x 10^–5 x 3 x 0.7314
= 89.96 \(\mu \mathrm{Wb}\).

Using right hand rule, current flows upwards.

R₀ = 3 Ω, T = 100^oC, T₀ = 20^oC
α = 0.004/^oC, R_T = ?
R_T = R₀(1 + α(T - T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
R₁₀₀ = 3.96 Ω

B = ∆m x c²
∆m = \(\frac{B}{c^2}\)
N M_n + Z M_p - M(N,Z) = \(\frac{B}{c^2}\)
N (N,Z) = N M_n + ZM_p - \(\frac{B}{c^2}\)

\(E=\frac{12400 \stackrel{o}A}{4100 \stackrel{o}A}=3.02 eV\)

\(\lambda\propto \frac{1}{\sqrt{V}}\)
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{224\times10^3}}{\sqrt{14\times 10^3}}\)
\(=\sqrt{16}=4\)
\(\lambda_{\mathrm{2}}= \frac{\lambda_1}{4}\)

d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)

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

\(L=\frac{20}{\pi^2} \mathrm{H}, \mathrm{f}=50 \mathrm{~Hz}  \)
\(f=\frac{1}{2 \pi \sqrt{L C}}  \)
\(50=\frac{1}{2 \pi \sqrt{\frac{20}{\pi^2} \times C}}  \)
\(50=\frac{1}{2 \times \sqrt{20 C}}  \)
\(\therefore(50)^2=\frac{1}{4 \times 20 C}  \)
\(\therefore C=\frac{1}{2500 \times 4 \times 20}=5 \times 10^{-6}=5 \mu \mathrm{F}\)

\(\phi_B =10 t^2-50 t+250  \)
\(e  =\frac{-d \phi_B}{d t}  \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50) 
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V

E = E_o sin(kx – ωt)
ω = ck
ω = 3 x 10⁸ x 10⁶
ω = 3 x 10¹⁴  rad s^-1 

Q = q₁ + q₂ = 4 x 10^-2C
\(q_{2f}=Q[\frac{r_2}{r_1+r_2}]\)
= 4 x 10^-2 \([\frac{3}{4}]\)
q_2f = 3 x 10^-2 C

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

Force ∝ charge 
After the separation, the magnitude of charge will be increased. So the force will be more than before.

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

R_s = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
R_s = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω

Given data:
In first case, f_e.= 0.03 m
f₀= 0.04 m , L = 0.2 m
\(m=\cfrac { L }{ { f }_{ o } } \left( 1+\cfrac { 0.25 }{ { f }_{ o } } \right) \) 
For m to be the same in both cases,
= \(\cfrac { L }{ 0.03 } \left( 1+\cfrac { 0.25 }{ 0.04 } \right) \) 
= \(\cfrac { 0.2 }{ 0.04 } \left( 1+\cfrac { 0.25 }{ 0.04 } \right) \) 
= \(\cfrac { L }{ 0.03 } \times \cfrac { 29 }{ 4 } =\cfrac { 0.2 }{ 0.04 } \times \cfrac { 28 }{ 3 } \) 
\(L=\cfrac { 5.6 }{ 29 } =0.193m\)

Path difference = \(\cfrac { xd }{ D } +\left( { n }_{ 2 }-{ n }_{ 1 } \right) t\) 
For P, = 0; x path difference
= (n² - n¹) t = 0.3 t
\(I={ I }_{ o }{ cos }^{ 2 }\cfrac { \phi }{ 2 } \) 
\(\cfrac { I }{ { I }_{ o } } ={ cos }^{ 2 }\cfrac { \phi }{ 2 } ,cos\cfrac { \phi }{ 2 } =\cfrac { \sqrt { 3 } }{ 2 } \) 
\(\cfrac { \phi }{ 2 } =\cfrac { \pi }{ 6 } ,\phi =\cfrac { \pi }{ 3 } \) 
Path difference = \(\cfrac { \lambda }{ 2\pi } \phi =\cfrac { 2\lambda }{ 2\pi } \cfrac { \pi }{ 3 } =\cfrac { \pi }{ 6 } \) 
\(0.3t=5\lambda +\cfrac { \pi }{ 6 } \)

The magnification due to the objective lens
\({ m }_{ o }=\cfrac { { v }_{ o } }{ \left( -{ \mu }_{ o} \right) } \) 
If the object is close to the focus of the objective lens then
u_o = f_o and v = L
(L = distance between two lenses)
\({ m }_{ o }=\cfrac { L }{ { f }_{ o } } \) 
If the final image is at the near point, then magnification due to the eye lens is
\({ m }_{ e }=\left( 1+\cfrac { D }{ { f }_{ e } } \right) \) 
\(M={ m }_{ o }\times { m }_{ e }=\cfrac { L }{ { f }_{ o } } \left( 1+\cfrac { D }{ { f }_{ e } } \right) \) 
The separation between the two lenses is 6.25 cm.

α = 0.98
I_Cu = 0.6 μA
\(\alpha =\frac { \beta }{ 1+\beta } \)
(or)
\(0.98=\frac { 1 }{ \frac { 1 }{ \beta } +1 } \)
β = 49
I_CEO = (1 + β)I_CBO
= (1 + 49) x 0.6μA
I_CEO = 30 μA
= (1 + 49) x 0.6 μA
I_CEO = 30 μA
I_C = βI_B+ I_CEO
= 49 x 20 μA + 30 μA
I_C = 1.01 mA

(i) This gives the variation of collector current (I_C) with changes in base current (I_B) at constant collector-emitter voltage (V_CE).
(ii) It is seen that a small Ie flows even when I_B is zero. This current is called the common emitter leakage current (I_CEQ) which is due to the flow of minority charge carriers.
Forward current gain:
(i) The ratio of the change in collector current (ΔI_C) to the change in base current (ΔI_B) at constant collector-emitter voltage (V_CE) is called forward current gain(β)
\(\beta ={ \left( \frac { \triangle { I }_{ C } }{ \triangle { I }_{ B } } \right) }_{ { V }_{ CE } }\)
(ii) It is value is very high and it generally ranges from 50 to 200. It depends on the construction of the transistors and will be provided by the manufacturer.

Principle : When a current carrying loop is placed in a uniform magnetic field it experiences a torque.
Construction : A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil as shown in Figure. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet.

The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. In a fine suspension strip, a small plane mirror is attached in order to measure the deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head. In order to pass electric current through the galvanometer, the suspension strip and the spring S are connected to terminals.
Working : Consider a single turn of the rectangular coil PQRS whose length be l and breadth b. PQ = RS = l and QR = SP = b.
Let I be the electric current flowing through the rectangular coil PQRS as shown in Figure. The horse-shoe magnet has hemi - spherical magnetic poles which produces a radial magnetic field. Due to this radial field, the sides QR and SP are always parallel to the magnetic field B and experience no force. The sides PQ and RS are always parallel to the magnetic field and experience force in opposite directions. Due to this, torque is produced.

For single turn, the deflection torque is,
て = bF = bBIl = (lb)BI
て = ABI
since, area of the coil A = lb
For coil with N turns, we get
て = NABI               ........(1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the moment of restoring couple is proportional to the amount of twist θ. Thus
て = Kθ      ............(2)
where K is the restoring couple per unit twist or torsional constant of the spring.
At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equations (1) and (2), we get,
NABI = Kθ
⇒ I =\(\frac { K }{ NAB } \) θ      ...........(3)
(or) I = Gθ
where G = \(\frac { K }{ NAB } \) is called galvanometer constant or current reduction factor of the galvanometer.
Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galvanometer we use are pointer type moving coil galvanometer.

In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)