Substances having (small) negative value (-0.5) of magnetic susceptibility \(\chi_m\) are diamagnetic.

Ratio = \(\frac{B_{1}}{\sqrt{B_{1}^{2}+B_{2}^{2}}}=\frac{1}{\sqrt{2}}\)

\(\frac{\Delta C}{C}=\frac{C^{\prime}-C}{C}=\frac{K C-C}{C}\)

(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.

(i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes the increasing flux.
(ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux.
(iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux. Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field

The insulating porcelain of the spark plugs accumulates a film of dirt.
The surface dirt is hygroscopic and picks up moisture from the air. Therefore, in humid weather, the insulating porcelain of the plugs becomes quasi-conductor. This allows an appreciable proportion of the spark to leak across the surface of the plug instead of discharging across the gap.

(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b) The tangent to a line of electric field at any point gives the direction of the electric field at that point. If any two lines of electric field cross each other, then at the intersection point, there would be two tangents and hence two directions for electric field, which is not possible. Hence, the electric field lines do not cross each other

Applying Kirchhoff's loop rule for loop ABEFA
6 + 3 + R₁ \(\times\) 0 - 3I₁ + 1 - 2I₁ = 0
or  10 - 5I₁ = 0
or I₁ = 2A
For loop BCDEB,
4 - I₁ . R + R₁  \(\times\) 0 - 3 = 0
or  1 - 2R = 0
\(\therefore \quad R=\frac{1}{2} \Omega\)

Potential difference between A and D through path
ABCD is 6 + 4 - I₁ R = V_AD \(\Rightarrow \quad 10-2 \times \frac{1}{2}=V_{A D}\)
\(\therefore\) V_AD = 9 V

Resistance of 50 watt bulb is more than that of 100 watt bulb. When 100 watt bulb is replaced by 50 watt bulb, connected in series with heater, the resistance of circuit increases and hence current decreases. \(H \propto I^2\) As , therefore the heater will now give less heat.

Given, total number of turns, N = 500
Length of solenoid, l = 0.5 m
Current, I = 5 A
Radius, r = 1 cm = 10^-2 m
Here, \(\begin{aligned} \frac{l}{r} & =\frac{0.5}{10^{-2}}=50 \Rightarrow l>>r \\ \end{aligned}\)
\(\begin{aligned} \therefore B & =\mu_0 n I=\frac{\mu_0 N I}{l} \\ \end{aligned}\)
\(\begin{aligned} =4 \pi \times 10^{-7} \times \frac{500}{0.5} \times 5 \end{aligned}\)
= 6.28 \(\times\) 10^-3 T

(a) When a charge +q is held at the centre of the shell, it will induce a charge -q on the inner surface of the shell of radius r and the charge +q on the outer surface of the outer shell.
Surface charge density on the inner surface of shell, 
As total charge on the outer surface of the outer shell = Q + q

There are three capacitors cach of capacitance 9 pF.
\(\therefore\) C₁ = C₂ = C₃ = 9 pF
and voltage, V = 120 V
(i) The total capcitance in series combination,
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
\(\Rightarrow \frac{1}{C_{s}}=\frac{3}{9} \Rightarrow C_{s}=3 pF\)
(ii) Let the charge across the system be q and potentials across C₁, C₂ and C₃ be V₁, V₂ and V₃, respectively.
Charge, q = C_s. V = 3 \(\times\)120 = 360 pC
Potential difference across C₁,
\(V_{1}=\frac{q}{C_{1}}=\frac{360}{9}=40 V\)
Potential difference across C₂,
\(V_{2}=\frac{q}{C_{2}}=\frac{360}{9}=40 V\)
Potential difference across C₃,
\(V_{3}=\frac{q}{C_{3}}=\frac{360}{9}=40 V\)
Thus, the potential difference across each capacitor is 40 V.

Given: \(q_{1}=4 \mu \mathrm{C}=4 \times 10^{-6} \mathrm{C}\)
\(q_{2}=-6 \mu \mathrm{C}=-6 \times 10^{-6} \mathrm{C}, r=20 \mathrm{~cm}\)
Let the electric potential be zero at a point P, a distance x (in cm) from q₁ Then
\(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{x}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{(r-x)}=0\)
\( \therefore \frac{4}{x} =-\frac{(-6)}{(20-x)} \)
\(\Rightarrow 4(20-x=6 x \)
\(x =8 \mathrm{~cm}\)
i.e. 8 cm from 4 μC charge.

\(60\mu F\)

1.2\(\Omega\)

Eddy currents: The induced circulating current produced in bulk piece of conductor, when subjected to a changing magnetic flux is called eddy current.
Applications:
(i) Magnetic braking in trains: In some electrically powered trains, strong electromagnets are situated above the rails. When these magnets are activated, eddy currents are induced in the rails. These currents oppose the motion of the train. Due to the absence of mechanical linkage, the breaking effect is strong.
(ii) Electromagnetic damping: Certain galvanometers have fixed core made of non-magnet metallic materials. When the coil of galvanometer oscillates, eddy currents generated on the core oppose its motion. As a result, the coil comes to rest quickly.
(iii) Electric power meters: The shiny metal disc in the electric power meter rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying current (i.e. ac) in the coil.

(a) Here, H = I = 5000 x 2 = 10⁴ A/m
and  I = XH
\(\therefore\) \(\chi=\frac{I}{H}\) 
\(=\frac{2 \times 10^{-2}}{10^{4}}=2 \times 10^{-6}\)
(b) According to Curie law.
\(x=\frac{c}{T}\)
\(\Rightarrow \quad \frac{\chi_{2}}{\chi_{1}}=\frac{T_{2}}{T_{1}}\)
\(\chi_{2}=\frac{T_{2}}{T_{1}} \chi_{1}=\frac{320}{300} \times 2 \times 10^{-6}\)
= 2.13 x 10^-6
\(\therefore\) Magnetisation at 320 K,
I = X₂H = 2.13 x 10^-6 x 10⁴
= 2.13 x 10^-2A/m

(ii) (a) Perpendicular to the plane of the paper and directed inward. 
(b) When angle between area vector of coil and magnetic field is 90°, then maximum torque experienced by the coil. When \(\theta\) = 0° or 180°, then torque will be minimum, i.e. zero.

\(Here,\ A=20\times 30cm^{ 2 }\)
\(=600\times { 10 }^{ -4 }{ m }^{ 2 }=6\times { 10 }^{ -2 }{ m }^{ 2 }\ B=0.3T\)
\({ \theta }_{ 1 }=\left( { 90 }^{ \circ }-{ 30 }^{ \circ } \right) =60^{ \circ };\quad { \theta }_{ 2 }={ 90 }^{ \circ }-{ 0 }^{ \circ }={ 90 }^{ \circ }\)
\({ \phi }_{ 1 }=BA \ cos{ \theta }_{ 1 }=0.3\times 6\times { 10 }^{ -2 }cos \ { 60 }^{ \circ } \ =0.9\times { 10 }^{ -2 }Wb\)
\({ \phi }_{ 2 }=BA \ cos \ { \theta }_{ 2 }=0.3\times 6\times { 10 }^{ -2 }cos \ { 90 }^{ \circ }=Zero\)

(a) 1.0 J, yes
(b) \(\omega \) = 10³ rad s^-1, v = 159 Hz
(c) 1.0 J