The properties of superconductor magnets are:
(i) perfect conductivity, and
(ii) perfect diamagnetism
\({ (i.e., } \left.\chi_{m}=-1, \mu_{r}=0\right)\)
Superconducting magnets are used for running magnetically levitated superfast trains.

Given, velocity of straight wire,
v = 5m/s
Horizontal component of the earth's magnetic field,

Length of the wire, \(\mathrm{I}=10 \mathrm{~m}\)
Falling speed of the wire, \(v=5.0 \mathrm{~m} / \mathrm{s}\)
Magnetic field strength, \(B=0.3 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
(a) Emf induced in the wire,
\( e=B l v \)
\(=0.3 \times 10^{-4} \times 5 \times 10 =1.5 \times 10^{-3} V\)
(b) Using Fleming's right-hand rule, it can be inferred that the direction of the induced emf is from West to East.
(c) The eastern end of the wire is at a higher potential.

Here, E = 9 \(\times\)10⁴ N/C, r = 2 cm = 2 \(\times\)10^-2 m and \(\lambda\)= ?
As, \(E=\frac{\lambda}{2\pi \varepsilon _{0}r}\Rightarrow \lambda =2\pi \varepsilon _{0}rE\)
\(=\frac{1}{2\times 9\times 10^{9}}\times 2\times 10^{-2}\times 9\times 10^{4}=10^{-7}Cm^{-1}\)

Number of turns in the solenoid, n = 800
Area of cross-section, A =  2.5 × 10^-4 m² 
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A
= 800 x 3 x 2.5 x 10^-4
= 0.6 J T^-1

Here, \(I_{ v }=1.5 \ mA=1.5\times 10^{ -3 }A,\)
\(\omega =300 \ rad/s, \ R=10 \ k\Omega =10^{ 4 }\Omega\)
\(C=0.5\mu F=0.5\times 10^{ -6 }F\)
\(X_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 300\times 0.5\times 10^{ -6 } } =\frac { 10^{ 6 } }{ 150 } \)
\(=\frac { 20 }{ 3 } \times 10^{ 3 }\Omega\)
\(Z=\sqrt { R^{ 2 }+X^{ 2 }_{ C } } =\sqrt { 10^{ 8 }+\frac { 4 }{ 9 } \times 10^{ 8 } } \)
\(=10^{ 4 }\sqrt { \frac { 13 }{ 9 } } =1.20\times 10^{ 4 }\Omega \)
rms voltage across the capacitor 
\(=I_{ v }X_{ C }=1.5\times 10^{ -3 }\times \frac { 20 }{ 3 } \times 10^{ 3 }=10volt\)

\(Here,\upsilon =500m/s \ \ e=? \ l=25m\)
\(R=5 \times { 10 }^{ -4 }T, \ \delta ={ 30 }^{ \circ }\)
\(e=Bl\upsilon =Vl\upsilon =(R \ sin \ \delta )l\upsilon \)
\(=5 \times { 10 }^{ -4 }sin{ 30 }^{ \circ } \times 25 \times 500=3.125 \ V\)

In the inductive circuit,
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos Φ
Where,
Φ = Phase difference between V and I
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
Hence, the net power absorbed can ve obtained as:
P = VI Cos Φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.

Given, inductance, L = 44 mH = 44 x 10^-3H, V_rms = 220V
Frequency of inductor, V = 50 Hz
Inductive reactance, X_L = 2\(\pi\)VL
= 2 x 3.14 x 50 x 44 x I0^-3 = 13.82 \(\Omega\)
The rms value of current in the circuit,
\(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_{L}}=\frac{220}{13.82}=15.9 \mathrm{~A}\)
Power absorbed, F = V_rms I_rms cos Φ
For pure inductive circuit, Φ = 90\(\unicode{xb0} \)
\(\therefore\) P = 0
Thus, power spent in one half cycle is retrieved in the other half cycle.

Here v = 0.5 Hz; N = 100, A = 0.1 m² and B = 0.01 T. Employing Equation.
\(\varepsilon_0=N B A(2 \pi v)\)
= 100 \(\times\)0.01 \(\times\)0.1 \(\times\) 2 \(\times\) 3.14 \(\times\) 0.5
= 0.314 V
The maximum voltage is 0.314 V.
We urge you to explore such alternative possibilities for power generation.

(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

Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T =  4.5 x 10^-2 J
The angle between the bar magnet and the external magnetic field, θ = 30°
Torque is related to magnetic moment (M) as:
\(T=M B \sin \theta \therefore M=\frac{T}{B \sin \theta}\)
\(=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 J T^{-1}\)
Hence, the magnetic moment of the magnet is 0.36  J T^-1.

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

a) Given: The peak voltage of supply is 100 V.
The rms voltage is give as,
v m = 2 ×V
Where, the peak value of supply voltage is v m and its rms value is V.
By substituting the given values in the above equation, we get
300= 2 ×V V= 300 2 =212.1 V
Thus, the value of rms voltage is 212.1V.
b) Given: The rms current in an ac circuit is 10 A.
The peak current in the circuit is given as,i m = 2 ×I
Where, the peak current in an ac circuit is i m and its rms value is I.
By substituting the given values in the above equation, we get
i m = 2 ×10 =14.1 A
Thus, the value of peak current in the given ac circuit is 14.1 A.

Method I
As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
dε = Bv dr . Hence,
\(\varepsilon = \int \mathrm{~d} \varepsilon=\int_0^R B v \mathrm{~d} r=\int_0^R B \omega r \mathrm{~d} r=\frac{B \omega R^2}{2}\)
Note that we have used v = ω r. This gives
\(\varepsilon=\frac{1}{2} \times 1.0 \times 2 \pi \times 50 \times\left(1^{2}\right)\)
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OP\(\mathcal{Q}\) in which point O and P are connected with a resistor R and O\(\mathcal{Q}\) is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B x (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OP\(\mathcal{Q}\) is given by
\(\pi R^{2} \times \frac{\theta}{2 \pi}=\frac{1}{2} R^{2} \theta\)
where R is the radius of the circle. Hence, the induced emf is
\(\varepsilon=B \times \frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2} R^{2} \theta\right]=\frac{1}{2} B R^{2} \frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{B \omega R^{2}}{2}\)
\(\text { [Note: } \left.\frac{\mathrm{d} \theta}{\mathrm{d} t}=\omega=2 \pi v\right]\)
This expression is identical to the expression obtained by Method I and we get the same value of ε.

Given, \(R=\frac{D}{2}=\frac{2.4}{2}=1.2 m\)
and \(\sigma =80.0 \quad \mu C/m^{2}\)
(i) Charge on sphere, q = 4\(\pi\)R² . \(\sigma\)
\(=4\times 3.14\times (1.2)^{2}\times 80=1446.912 \quad \mu C\)
=1.45 \(\times\)10^-3C
(ii) Electric flux, \(\phi=\frac{q}{\varepsilon_{0}}=\frac{1.45\times 10^{-3}}{8.854\times 10^{-12}}\)
= 0.1637 \(\times\)10⁹ = 1.637 \(\times\)10⁸ N-m²/C

(i) The phase difference berween E and I is \(\frac{\pi}{2}\). the voltage E leads the current I by phase angle \(\frac{\pi}{2}\).
(ii) The phasor diagram is given by

(iii) L = 3 H, E_rms = 150 V, f = 50 Hz
\(I_0=\frac{E_0}{X_L}=\frac{\sqrt{2} E_{\mathrm{mms}}}{2 \pi f L}=\frac{\sqrt{2} \times 150}{2 \pi \times 50 \times 3}=0.225 \mathrm{~A}\)
(iv) f = 200 Hz, L = 15 mH = 15 \(\times\)10^-3 H
\(\therefore\) Inductive reactance, X_L = \(\omega\)L = 2\(\pi\)fL
= 2 \(\pi\) \(\times\)200 \(\times\)15 \(\times\)10^-3 = 18.84 \(\Omega\)

(i) Dipole moment of bar magnet = M = m . l
When it is bent into semi-circular wire then,

Radius of semi-circle, \(r=\frac{l}{\pi}\)
\(\therefore\) Distance between the two poles is l' = 2r = \(\frac{2l}{\pi}\)
\(\therefore\) Dipole moment of semi-circle is \(M^{\prime}=m \cdot l^{\prime}=\frac{2 m l}{\pi}=\frac{2 M}{\pi}\)
(ii) The dipole moment vector for the arrangement is shown below,

\(\therefore\) The resultant dipole moment is given by
\(M^{\prime}=\sqrt{M^2+M^2}=\sqrt{2} M\)
(iii) The direction of magnetic dipole moment is from south pole to north pole of a magnetic dipole.
(iv) Unlike a electric charge, an isolated magnetic pole (either north or south pole) do not exist.