Given, magnetic moment of magnet, M = 0.32 J/T
Magnitude of magnetic field, B = 0.15 T
(i) For stable equilibrium, the angle between magnetic moment M and magnetic field B is \(\theta\) = 0°.
[\(\because\) In this position, it will be in a direction parallel to the magnetic field, thus no torque will act on it.]
\(\because\) The potential energy of the magnet,
U = - M . B = -MB cos \(\theta\)   [\(\because\) M.B = MB cos \(\theta\)]
=-0.32 \(\times\)0.15 cos 0° = -4.8 \(\times\)10^-2 J
Thus, for the stable equilibrium the potential energy is -4.8 \(\times\)10^-2 J.
(ii) For unstable equilibrium, the angle between the magnetic moment and magnetic field is 180°.
(\(\because\)At \(\theta\)=180°, although torque is zero but if it is displaced by small angle d\(\theta\), then resulting torque would not restore it to the original position).
Potential energy of the magnet,
U = - MB cos 180° = -0.32 \(\times\)0.15 (-1)
= 4.8 \(\times\)10^-2 J
Thus, for the unstable equilibrium, the potential energy is 4.8 \(\times\)10^-2 J.

Working Let initially positively charged is accelerated towards D₂ and enter into it.
Now, the charged particle experiences magnetic Lorentz force due to a strong normal magnetic field. It performs circular motion. The time taken by the charge particle to complete half revolution is equal to half of time period of AC oscillator between two dees.
The charge d particle again accelerated towards D₁ as D₂ acquires positive and d negative polarity. Thus, the charge particle is brought again and again in the small region of oscillating electrical field by strong normal magnetic field.
In case of the cyclotron, the particle moves on a circular path, the centripetal force required is provided by magnetic force, so magnetic
Lorentz force = centripetal force
qvB = \(\frac{mv^{2}}{r} \Rightarrow r=\frac{mv}{qB}\Rightarrow v=\frac{qBr}{m}\)
\(\therefore KE=\frac{1}{2}mv^{2}=\frac{1}{2}m(\frac{qBr}{m})^{2}=\frac{q^{2}B^{2}r^{2}}{2m}\)
For maximum KE, r = r_o (radius of dees).
(ii) Given, I₁ = 2A, I₂ = 5A, r = 1 cm = 1 x 10^-2m
Force per unit length between two wires.
\(\frac{F}{L}=\frac{\mu_0}{4\pi}\frac{2I_1 I_2}{r}=\frac{10^{-7}\times2\times2\times5}{1\times10^{2}}=20\times10^{-5}\)
\(\frac{F}{L}=2\times10^{4}Nm^{-1}\)
Currents flowing in wires are in opposite directions, so the force will be of repulsive nature.

Magnetic field strength, B = 6.5 G = 6.5 x 10^–4 T
Speed of the electron, v = 4.8 x 10⁶ m/s
Charge on the electron, e = 1.6 x 10^–19 C
Mass of the electron, m_e = 9.1 x 10^–31 kg
Angle between the shot electron and magnetic field, θ = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sinθ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
\(F_{c}=\frac{m v^{2}}{r}\)
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
F_C = F
\(\frac{m v^{2}}{r}=e v B \sin \theta\)
\(r=\frac{m v}{B e \sin \theta}\)
\(=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^{6}}{6.5 \times 10^{-4} \times 1.6 \times 10^{-19} \times \sin 90^{\circ}}\)
= 4.2 x 10 ^-2 m = 4.2 cm
Hence, the radius of the circular orbit of the electron is 4.2 cm.

Magnetic permeability: It is a measure of the extent to which magnetic field influence can pass through a material.
The following are the characteristics:
(i) Material should have high retentivity.
(ii) Material should have high coercivity.
(iii) Material should have high permeability.
A suitable material for permanent magnet is alnico steel.

Since, \(L_{N_{1} S_{2}}=\sqrt{L^{2}+L^{2}}=\sqrt{2} L\) 
So, the magnetic moment of the system is calculated as
\(M=m \times \sqrt{2} L=\sqrt{2} m L\)

Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
∴  Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, B \(=\frac{\mu_{0} 2 I}{4 \pi r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 50}{4 \pi \times 2.5}\)
= 4 x 10 ^-6 T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Magnitude of one of the magnetic fields, B₁ = 1.2 x 10⁻² T
Angle between the two fields, θ = 60°
At stable equilibrium, the angle between the dipole and field B₁, θ₁ = 15°
Let the magnitude of the other the other magnetic field be B₂
Angle between the dipole and field B₂, θ₂ = θ − θ₁ = 60° − 15° = 45°
At rotational equilibrium, the torques due to both the fields must balance each other. That is,
Torque due to field B1 = Torque due to field B2Torque due to field B1 = Torque due to field B2
⇒ MB1 sin θ₁ = MB2  sin θ2 
\(\Rightarrow \mathrm{B}_{2}=\frac{\mathrm{B}_{1} \times \sin \theta_{1}}{\sin \theta_{2}}\)
Where,
M = Magnetic moment of the dipole
Putting the values, we get,
\(\mathrm{B}_{2}=\frac{1.2 \times 10^{-2} \times \sin 15^{\circ}}{\sin 45^{\circ}}\)
= 4.39 x 10^-3 T
Therefore, the magnitude of the other magnetic field is 4.39 x 10⁻³ T.

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, B = 0.02 T, r = 6 x 10^-2 m;
pitch = 24 x 10^-2 m
Let V₁ be the component of velocity of proton perpendicular to magnetic field. Then
 \(r=\frac { { mv }_{ 1 } }{ qB } \ or \ { V }_{ 1 }=\frac { rqB }{ m } \) 
\(\therefore { V }_{ 1 }=\frac { \left( 6\times { 10 }^{ -2 } \right) \times \left( 1.6\times { 10 }^{ -19 } \right) \times 0.02 }{ 1.6\times { 10 }^{ -27 } } \) 
= 1.2 x 10⁵ ms^-1
Time period of revolution,
 \(T=\frac { 2\pi r }{ { v }_{ 1 } } =\frac { 2\pi \times 6\times { 10 }^{ -2 } }{ 1.2\times { 10 }^{ 5 } } =\pi \times { 10 }^{ -6 }s\) 
Let V₂ be the component of velocity of proton along the direction of magnetic field. Then 
 \({ V }_{ 2 }=\frac { pitch }{ time\quad period } =\frac { 24\times { 10 }^{ -2 } }{ \pi \times { 10 }^{ -6 } } \) 
= 7.6 x 10⁴ ms^-1

If \({ i }_{ a }\)and \({ i }_{ b }\) are the 3 antiparallel currents flowing in two parallel wires separated by a distance R, then the repulsive force experienced by unit length (1m) of either wire is
\(F=\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi R } \)
Weight per meter length of above wire
\(=0.073{ Nm }^{ -1 }\)
According to problem, the weight of lower wire is supported by upward magnetic repulsion, therefore
\(\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi R } \)\(=0.073\)
\(R=\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi \times 0.073 } \) 
\(=\frac { 4\pi \times { 10 }^{ -7 }\times 100\times 20 }{ 2\pi \times 0.073 } \)
\(=4.4\times { 10 }^{ -3 }m = 4.48mm\)

The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v x B is along the z-axis (screw rule or right-hand thumb rule). So,
(a) for electron it will be along –z axis.
(b) for a positive charge (proton) the force is along + z axis.

As we know that, the magnetic moment of electron or proton is inversely proportional to the mass of electron or proton, respectively.
Magnetic. moment of electron, \(\mu_{e} \text { oc } \frac{1}{m_{e}}\)
where, m_e = mass of electron
Magnetic moment of proton, \(\mu_{p} \propto \frac{1}{m_{p}}\)
where, m_p = mass of proton
\(\frac{\mu_{e}}{\mu_{p}}=\frac{m_{p}}{m_{e}}\)  .......(i)
As, we know that, m_e < < m_p
So, \(\frac{m_{p}}{m_{e}}>>1 \Rightarrow \frac{\mu_{e}}{\mu_{p}}>>1\)
\(\Rightarrow\) \(\mu_{e}>>\mu_{p}\)
Thus, as the value of magnetic moment of electron is much more as compared to magnetic moment of proton, so the effect of proton is neglected.

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.

\({ N }_{ 1 }.2\pi R={ N }_{ 2 }.2\pi (R/2)\)
  \(\therefore \ { N }_{ 2 }=2{ N }_{ 1 }\)
Magnetic moment of a coil, M =NAI
For the coil of radius 'R'
  \({ M }_{ 1 }={ N }_{ 1 }{ IA }_{ 1 }={ N }_{ 1 }I\pi { R }^{ 2 }\)
For the coil of radius R/2
\({ M }_{ 2 }={ N }_{ 2 }{ IA }_{ 2 }=2{ N }_{ 1 }I\pi { R }^{ 2 }/4={ N }_{ 1 }.\pi { R }^{ 2 }/2\)
 \({ M }_{ 1 }:{ M }_{ 2 }=1:2\) 

M = NIA, when N is number of turns I, the current and A = area of cross section of loop.

(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.

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(b): There will be only one neutral point on the horizontal board. This is because field of earth magnetic field is from south to north; and the field of pole on the board is radially outwards. At any point towards south of magnetic pole, field of earth and field of pole will cancel out to give a neutral point.

(d): In a non-uniform magnetic field, a torque and a net force both act on the dipole. If magnetic field is uniform, net force on dipole would be zero.