In this situation, the majority charge carriers will flow in the emitter-base circuit and also in the collector-base circuit, In this case, the working of transistor will be equivalent to two \(p-n\) junction diodes with a common base. It means the purpose of transistor will be defeated.

Here, N = 300, \(r=\frac { 14 }{ 2 } \) = 7 cm = 7 x 10^-2 m, 
I = 15 A
\(M=NIA=NI\left( \pi { r }^{ 2 } \right) \) 
\(=300\times 15\times \frac { 22 }{ 7 } \times { \left( 7\times { 10 }^{ -2 } \right) }^{ 2 }\) 
M = 69.3 JT^-1

The magnetic field in the empty space enclosed by the windings of the toroid is given by
\(B={ \mu }_{ 0 }nI\)
where n is the number of turns per unit length and I is the current.If space is filled by a core of permeability \(\mu \)the formula modifies to
\( d B=\mu nI\)
\(\\ here \ B=2.5T,I=0.70A\)
\(n=\frac { 3000 }{ 2\pi \times 11.5\times { 10 }^{ -5 } } { m }^{ -1 }\ \)
where we have ignored the variation os B across the cross-section of the toroid and taken the radius of the toroid to be the mean of inner and outer radii.We then have
\(\mu =\frac { B }{ nI } =8.61\times { 10 }^{ -4 }{ TmA }^{ -1 }\quad \)
The relative permeability of the core is defined to be \(\frac { \mu }{ { \mu }_{ 0 } } \) 
with  \({ \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }{ TmA }^{ -1 }\)
we get  \(\frac { \mu }{ { \mu }_{ 0 } } =685\) 

There is no contradiction. Field lines inside the bar go away from S towards N. The next flux of B over any surface fully enclosing N or S must be identically zero. 

Number density of free electrons in a copper conductor, n = 8.5 x 10²⁸ m^-3 Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10^-6 m²
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeV_d
Where,
e = Electric charge = 1.6 x 10⁻¹⁹ C
V_d = Drift velocity = \(\frac{\text { Length of the wire (l) }}{\text { Time taken to cover l(t) }}\)
\(I=n A e \frac{l}{t}\)
\(t=n A e \frac{l}{I}\)
\(=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
\(=2.7 \times 10^{4} s\)
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 10⁴ s.

ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge q =5 \(\mu\)C is placed. O is the centre of the hexagon.

Given, AB = BC = CD = DE
= EF = FA = d = 10 cm
As, the hexagon has six equilateral triangles, so the distance of centre O from every vertex is 10 cm.
i.e. OA = OB = OC = OD
= OE = OF = d = 10 cm
\(\therefore\) Potential at point O = Sum of potentials at centre O due to individual point charge
i.e. V_O = V_A + V_B + V_C + V_D + V_E + V_F
\(=\frac{1}{4\pi \varepsilon_{0}}.\left [ \frac{q}{OA}+\frac{q}{OB}+\frac{q}{OC}+\frac{q}{OD}+\frac{q}{OE}+\frac{q}{OF} \right ]\)
\(=\frac{1}{4\pi \varepsilon _{0}}.\frac{6q}{d}\)    \(\left [ \because V=\frac{1}{4\pi \varepsilon _{0}.\frac{q}{r}} \right ]\)
Putting the values, we get
\(=9\times 10^{9}\times \frac{6\times 5 \times 10^{-6}}{10\times 10^{-2}}\)
= 2.7 \(\times\)10⁶ V

Given, resistance of heating element at temperature 27°C,   
R₂₇ = 100 \(\Omega\)
Resistance of heating element at temperature t°C,
R_t = 117 \(\Omega\)
\(\alpha\)=1.70 \(\times\)10^-4 °C^-1, t = ?
By using the formula of temperature coefficient of resistance,
\(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\)   ...(i)
Here, R₁ = R_t, R₁ = R₂₇, t₂ = t and t₁ = 27°C
Such that, \(\alpha=\frac{R_t-R_{27}}{R_{27}(t-27)}\)
 Substituting given values in Eq. (i), we get
\(\begin{aligned} 1.70 \times 10^{-4} & =\frac{117-100}{100(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{17}{100 \times 1.70 \times 10^{-4}} \end{aligned}\)
or t = 1000 + 27 = 1027°C 

Length of the solenoid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
∴ Total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
\(B=\frac{\mu_{0} N I}{l}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}\)
\(=8 \pi \times 10^{-3}=2.512 \times 10^{-2} T\)
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 x 10^–2 T.

\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2} \times \frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 x 10⁴ V 
(b) W = qV = 2 x 10⁻⁹C x 4 x 10⁴ V
= 8 x 10^–5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Given, potential difference = 230 V
Initial current at 27°C = I_27°C = 3.2 A
Final current at t°C = I_t°C = 2.8 A
Room temperature = 27°C
Temperature coefficient of resistance, \(\alpha=1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
Resistance at 27°C, R_27°C\(=\frac{V}{I_{27^{\circ} \mathrm{C}}}=\frac{230}{3.2}=\frac{2300}{32} \Omega\)
Resistance at t°C, R_t°C = \(\frac{V}{I_{t^{\circ} \mathrm{C}}}=\frac{230}{2.8}=\frac{2300}{28} \Omega\)
Temperature coefficient of resistance
\(\begin{aligned} \alpha & =\frac{R_t-R_{27}}{R_{27}(t-27)} \end{aligned}\)
\(\begin{aligned} \Rightarrow 1.70 \times 10^{-4} & =\frac{\frac{2300}{28}-\frac{2300}{32}}{\frac{2300}{32}(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{82.143-71.875}{71.875 \times 1.70 \times 10^{-4}}=840.347 \end{aligned}\)
or t = 840.3 + 27 = 867.3 °C
Thus, the steady temperature of heating element is 867.3 °C

(a) Comparing the given equation with
\(B_{y}=B_{0} \sin \left[2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)\right]\)
We get, \(\lambda=\frac{2 \pi}{0.5 \times 10^{3}} \mathrm{~m}=1.26 \mathrm{~cm}\)
and \(\frac{1}{T}=v=\left(1.5 \times 10^{11}\right) / 2 \pi=23.9 \mathrm{GHz}\)
(b) E₀ = B₀c = 2 x 10^–7 T x 3 x 10₈ m/s = 6 x 10¹ V/m
The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as
E_z = 60 sin (0.5 x 10³x + 1.5 x 10¹¹ t) V/m.

(a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value.
(b) Each ‘free’ electron does accelerate, increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
(c) Simple, because the electron number density is enormous, ~10²⁹ m^-3.
(d) By no means. The drift velocity is superposed over the large random velocities of electrons.
(e) In the absence of electric field, the paths are straight lines; in the presence of electric field, the paths are, in general, curved.

Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
\(I=\frac{E}{r}\)
\(=\frac{12}{0.4}=30 A\)
The maximum current drawn from the given battery is 30 A.

(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus the frequency of scattered light equals the frequency of incident light.
(b) No, Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon crossing an unit area per uni time.

Given,
The area of plate of the capacitor, A = 6 x 10^-3 m²
Distances between the plates, d = 3mm = 3 x 10^-3 m
Voltage supplied, V = 100V
Capacitance of a parallel plate capacitor is given by, \(C=\frac{\epsilon \times A}{d}\) 
Here,
ε = permittivity of free space = 8.854 x10^-12 N^-1 m ^-2 C^-2
\(C=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}=17.81 \times 10^{-12} \mathrm{~F}=17.71 \mathrm{pF}\)
Therefore, each plate of the capacitor is having a charge of
q = VC = 100 x 17.81 x 10^-12 C = 1.771 x 10^-9 C

(i) (a) x-rays or y-rays
(b) x-rays
(c) Microwaves: Microwaves are used in RADAR for aircraft navigation/cooking food in microwave oven/knowing the speed of vehicles on road.
(ii) c = 3.0 x 10 ⁸ m/s
\(\lambda\) = 1.0 x 10^{- 12} m
\(\therefore \quad v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{10^{-12}}=3 \times 10^{20} \mathrm{~Hz}\)

(i) (c): In mass spectrometer, the ions are sorted out by accelerating them through electric and magnetic field.
(ii) (c): As \(\frac{m v^{2}}{r}=q v B_{0} \therefore r=\frac{m v}{q B_{0}}\)
(iii) (b): As radius \(r \propto \frac{m}{q}\)
\(\therefore\) r will be maximum for \(\alpha\) - particle.
(iv) (b) : Here, \(r=\frac{m v}{q B_{0}} \text { or } m=\frac{r q B_{0}}{v}\)
As \(v=\frac{E}{B}, \therefore m=\frac{q B_{0} B r}{E}\)
(v) (c): From the relation v = E/B, it is clear electric and magnetic force balance each other.