Given: v = 900 km h^-1 = 250 ms^-1; l = 20 m B_v (vertical component of B)
= 5 x 10^-4 sin 30°
= 2.5 x 10^-4 T [∵ s, = B sin \(\delta\)]
∴ Induced emf = e = B/v
= 2.5 x 10^-4 x 20 x 250 V
= 1.44 V

Length of the rod, l = 1 m
Angular frequency,ω = 400 rad/s
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.
Average linear velocity of the rod,

Emf developed between the centre and the ring,

Hence, the emf developed between the centre and the ring is 100 V.

Let I₁ and I₂ be the currents in two cells with emfs, E₁ and E₂ and internal resistances, r₁ and r₂

So, I = I₁ + I₂
Now, let V be the potential difference between the points, A and B. Since, the first cell is connected between the points A and B
V = potential difference across first cell
V = E₁-I₁r₁ or I₁=\(\frac { { E }_{ 1 }-V }{ { r }_{ 1 } } \)
Now, the second cell is also connected between the points, A and B. So,
I₂ = \(\frac { { E }_{ 2 }-V }{ { r }_{ 2 } } \)
Thus, substituting for I₁ and I₂
\(I=\frac { { E }_{ 1 }-V }{ { r }_{ 1 } } +\frac { { E }_{ 2 }-V }{ { r }_{ 2 } } \)
or \(I=\left( \frac { { E }_{ 1 } }{ { r }_{ 1 } } +\frac { E_{ 2 } }{ { r }_{ 2 } } \right) -V\left( \frac { 1 }{ { r }_{ 1 } } +\frac { 1 }{ { r }_{ 2 } } \right) \)
\(V=\left( \frac { { E }_{ 1 }{ r }_{ 2 }+{ E }_{ 2 }{ r }_{ 1 } }{ { r }_{ 1 }+{ r }_{ 2 } } \right) -I\left( \frac { { r }_{ 1 }{ r }_{ 2 } }{ { r }_{ 1 }+{ r }_{ 2 } } \right) \)            ........(i)
If E is effective emf and r, the effective internal resistance of the parallel combination of the two cells, then
V = E - Ir              ..............(ii)
Comparing Eqs. (i) and (ii), we get
(i) \(E=\frac { { E }_{ 1 }{ r }_{ 2 }+{ E }_{ 2 }{ r }_{ 1 } }{ { r }_{ 1 }+{ r }_{ 2 } } \)
This is equivalent emf of the combination.
(ii)\(\frac { { r }_{ 1 }{ r }_{ 2 } }{ { r }_{ 1 }+{ r }_{ 2 } } \)
This is equivalent resistance of the combination.
(iii) the potential difference between the points A and B is
V = E - Ir

Mutual inductance of a pair of coils, µ = 1.5 H
Initial current, I₁ = 0 A
Final current I₂ = 20 A
Change in current, 
Time taken for the change, t = 0.5 s
Induced emf, 
Where  is the change in the flux linkage with the coil.
Emf is related with mutual inductance as:

Equating equations (1) and (2), we get

Hence, the change in the flux linkage is 30 Wb.

The current in the conductor having length l₁ cross-sectional area A and number density n is
I = neAv_d          ...........(i)
Electric field inside the wire is given by
E = V/l         ...............(ii)
If relaxation time is ፒ, the drift speed
\({ v }_{ d }=\frac { e\tau E }{ m } \)
where, m = mass of electron
ፒ = relaxation charge
e = electronic charge
E = electric field.
\(I=\frac { ne^{ 2 }\tau }{ m } AE\)             .........(ii)
From Eqs. (ii) and (iii). we get
\(\frac { ne^{ 2 }\tau AV }{ ml } \Rightarrow J=\frac { 1 }{ A } =\frac { ne^{ 2 }\tau V }{ ml } \).

Relaxation time (\(\tau\) ): The average time interval between two successive collisions for the free electrons drifting within a conductor (due to the action of the applied electric field), is called relaxation time.
Relation
\(v_d=(-eE\tau)/m\)
Since \(i=-n\ e A\ v_d\)
\(=ne^2 \ A\ \tau\ V/ml\)
\(\therefore V/i=ml/(ne_2\ A\ \tau)=pl/A\)
\(\therefore p=m/(ne^2\tau )\)

Given  \(\phi =5{ t }^{ 3 }+4{ t }^{ 2 }+2t-5,\quad t=2s,\)
\(R=10\Omega \)
\(\therefore \)  Magnitude of induced e.m.f.
\(e=\left| e \right| =\frac { d\phi }{ dt } \)
or \(e=\frac { d }{ dt } \left[ 5{ t }^{ 3 }+4{ t }^{ 2 }+2t-5 \right] \)
\(=15{ t }^{ 2 }+8t+2\)
\(=15\times 4+8\times 2+2\)
or  \(e=78V\)
\(\therefore \)  \(I=\frac { e }{ R } =\frac { 78 }{ 10 } =7.8A\)

Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

(ii) To deduce the expression for the power of the combination, first find the equivalent resistance of the combination in the given conditions.
\(\because \ P_{1}=\frac{V^{2}}{R_{1}} \Rightarrow R_{1}=\frac{V^{2}}{P_{1}}\)
and \(P_{2}=\frac{V^{2}}{R_{2}} \Rightarrow R_{2}=\frac{V^{2}}{P_{2}}\)
(a) In series combination,
\(R_{s}=R_{1}+R_{2}=\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}\)
\(\Rightarrow \ R_{s}=V^{2}\left(\frac{1}{P_{1}}+\frac{1}{P_{2}}\right)=V^{2}\left(\frac{P_{1}+P_{2}}{P_{1} P_{2}}\right)\)
Now, let the power of heating element in series combination be P_s.
\(\therefore \ P_{s}=\frac{V^{2}}{R_{1}+R_{2}}=\frac{V^{2}}{V^{2}\left(\frac{P_{1}+P_{2}}{P_{1} P_{2}}\right)}=\frac{P_{1} P_{2}}{P_{1}+P_{2}}\)
(b) In parallel combination,
\(\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{\frac{V^{2}}{P_{1}}}+\frac{1}{\frac{V^{2}}{P_{2}}}=\frac{P_{1}}{V^{2}}+\frac{P_{2}}{V^{2}}\)
\(\Rightarrow \ \frac{1}{R_{p}}=\frac{1}{V^{2}}\left(P_{1}+P_{2}\right)\)
Now, power consumption in parallel combination,
\(P_{p}=\frac{V^{2}}{R_{p}}=V^{2}\left(\frac{1}{R_{p}}\right)\)
\(\Rightarrow \ P_{p}=V^{2}\left[\frac{1}{V^{2}}\left(P_{1}+P_{2}\right)\right]\)
∴ P_p = P₁ + P₂

Here, number of turns per unit length,
\(n=\frac{N}{l}=15\) turns/cm = 1500 turns/m
A = 2.0 cm² = 2 \(\times\)10^-4m²
\(\begin{aligned} \therefore \frac{d I}{d t}=\frac{4-2}{0.1} \text { or } \frac{d I}{d t}=20 \mathrm{As}^{-1} \\ \end{aligned}\)
\(\begin{aligned} \therefore|e|=\frac{d \phi}{d t}=\frac{d}{d t}(B A) \quad\left[\because B=\frac{\mu_0 N I}{l}\right] \\ \end{aligned}\)
\(= \frac{A d}{d t}\left(\mu_0 \frac{N I}{l}\right)=A \mu_0\left(\frac{N}{l}\right) \frac{d l}{d t}\)
= (2 \(\times\) 10^-4) \(\times\)4 \(\pi\)\(\times\)10^-7 \(\times\)1500 \(\times\)20 V
= 7.5 \(\times\)10^-6 V

(i) Mutual inductance, between the two coils is equals to the magnetic flux linked with one coil when a unit current is passed in the other coil.
Alternatively,
\(e=\frac { -MdI }{ dt } \)
Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.
SI unit: henry, (weber ampere^-1) or (volt second ampere^-1).
(ii)

Let a current I₂ flow through S₂. This sets up a magnetic flux ф₁ through each turn of the coil S₁. Total flux linked with S₁
 N₁ф₁= M₁₂ I₂   ...(i)
where M₁₂ is the mutual inductance between the two solenoids
Magnetic field due to the current I₂ in S₂ is μ₀n₂I₂.
Therefore, resulting flux linked with S₁
N₁ф₁=[(n₁l)\({ \pi r }_{ 1 }^{ 2 }\)] (μ₀n₂I₂)   ....(ii)
Comparing (i) and (ii), we get
M₁₂I₂ =(n₁l) (μ₀n₂I₂)
M₁₂ =μ₀n₂n₁ \({ \pi r }_{ 1 }^{ 2 }\)l
(iii) Let a magnetic flux be (ф₁) linked with coil C₁ due to current (I₂)in coil C₂.
We have
ф₁ ∝ I₂
⇒ ф₁ = MI₂
∴  \(\frac { d{ \phi }_{ 1 } }{ dt } =M\frac { { dI }_{ 2 } }{ dt } \)
⇒   e= -M \(\frac { { dI }_{ 2 } }{ dt } \)

It works on the principle of electromagnetic induction, i.e. when a coil continuously rotates in a magnetic field, the magnetic flux associated with it keeps on changing; thus an emf is induced in it.
(b) When the coil rotates in a magnetic field, its effective area i.e. A cos \(\theta \) , (i.e. area normal to the magnetic field) keeps on changing. Hence magnetic flux  \(\phi \) = NBAcos S, keeps on changing.
(c) Let the coil be rotating with angular velocity 'm',
at any instant 't' when the normal to the plane of the coil makes an angle \(\theta \) with the magnetic field. Hence magnetic flux, 12   \(\phi \) = NBA cos mt, Therefore induced emf (e)
\(\epsilon =-\frac { d\phi }{ dt } \)
\(\Rightarrow \epsilon =NBA\omega sin\quad \omega t\)
Induced emf will be maximum when mt = 90° 12 Hence, max = NBA\(\omega \)
Direction of induced emf can be determined using Flemming's Right-hand rule. Alternatively: Statement of the above rule.

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA } ....(1)\)
Also drift velocity in terms of average relaxation time τ is given by\(\)
\({ v }_{ d }=\frac { eE\tau }{ m } .......(2)\)
From (1) and (2), we have
\(\frac { eE\tau }{ m } =\frac { I }{ neA } \)
\(or \ \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or \ \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or \ \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau }........(3)\)
The R.H.S. of Eq(3) is constant
\(\frac { V }{ I } = \ Constant\)
This is Ohm′s law
\(But \ \frac { V }{ I } =R, \)the resistance of the conductorSo Eq(3) becomes
\(R=\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(But\quad R=\rho \frac { l }{ A }\)
\(\therefore \ \rho \frac { l }{ A } =\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\( \rho =\frac { m }{ n{ e }^{ 2 }A\tau } \)

Electromagnetic induction
Electromagnetic induction may be defined as the phenomenon of production of electric current (or e.m.f.) in a closed coil (or circuit), when magnetic flux linked with the coil (or circuit) is changed.
Magnetic flux \(\left( \phi \right) \). The total number of lines of induction passing normally through a given area is called magnetic flux. Consider a surface A. Take a small element of area dA. If \(\overrightarrow { B } \) makes an angle \(\theta \) with the area vector \(d\overrightarrow { A } \), then magnetic flux linked with area dA will be
\(d\phi =\overrightarrow { B } .d\overrightarrow { A } \)
(where area vector is a vector \(d\overrightarrow { A } \) having magnitude dA and direction normal to \(dA\))

Total magnitude flux linked with the whole surface S will be
\(\phi =\int { \overrightarrow { B } .d\overrightarrow { A } } \)
If the surface is plane and having area S, then
\(\phi =\overrightarrow { B } .\overrightarrow { A } \)
or  \(\phi =BA\cos { \theta } \)    .....(1)
Units of \(\phi \)
If \(B=1T,A=1{ m }^{ 2 },\theta ={ 0 }^{ ° },then\quad \phi =1Wb.\)
Hence magnetic flux is said to be 1 weber (1Wb) if a magnetic field of 1 tesla is normal to the area of 1 \({ m }^{ 2 }.\)
So  \(1Wb=1T{ m }^{ 2 }\)
Magnetic induction (B). It is defined as the magnetic flux associated per unit area.
i.e. \(B=\frac { \phi }{ A } \)
Units. In S.I. the unit of B is Wb \({ m }^{ 2 }\) or tesla (T).
Positive and negative flux
Since \(\phi =BA\cos { \theta } \)
So (I) \(If \ \theta >{ 90 }^{ ° },\phi =-ve\left( i.e. \ \phi \ is \ inward \ the \ surface \right) \)
(ii) \(If \ \theta <{ 90 }^{ ° },\phi =+ve\left( i.e. \ \phi \ is \ outward \ to \ the \ surface \right) \)
(iii) \(If \ \theta ={ 90 }^{ ° },\phi =BS\left( i.e. \ maximum \ flux \right) \)
Dimensional formula of \(\phi \)
\(\phi =BA\cos { \theta =\frac { FA }{ q\upsilon } \cos { \theta \left[ \therefore F=qB\upsilon \right] } } \)

\(=\frac { \left[ ML{ T }^{ -2 } \right] \left[ { L }^{ 2 } \right] }{ \left[ AT \right] \left[ L{ T }^{ -1 } \right] } \)

\(=\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -1 } \right] \)

(a): The coefficient of self inductance of the coil is given by \(L=\frac{\mu_{0} N^{2} A}{l}\)
where N is number of turns, 1 is length of the coil and
 A is area of coil, so \(L \propto N^{2}\)

(c): The manner in which the two coils are oriented, determines the coefficient of coupling between them \(\text { i.e., } K=\sqrt{\frac{M}{L_{1} L_{2}}}\), where L, and L2 are self-inductance of two coils. When the two coils are wound on each other, the coefficient of coupling is maximum and hence mutual inductance between the coil is maximum.

(d): When two bulbs are connected in series, the resistance of the circuit increases and so the voltage in each decreases, hence the brightness and the temperature also decreases. Due to decrease in temperature, the resistance of the carbon filament will slightly increase while that of metal filament will decrease. Hence, carbon filament bulb will glow more brightly (P = i²R). Also carbon is not a semiconductor

(a): In normal conductor, the direction of electrons are randomly oriented such that the total sum of their velocities is equal to zero.