Force on a moving charge in magnetic field is given as F = qvB sin \(\theta\)
When charge particle is moving along the direction of magnetic field, then
\(\theta\)= 0°  \(\Rightarrow\) F = 0

As we know that in a circular path, frequency of a charged particle is given by
\(v=\frac{q B}{2 \pi m} \text { or } \quad v \propto \frac{1}{m}\)
Since, m_p > m_e therefore electron will move in circular path with higher frequency.

Biot-Savart's law states that, the magnitude of magnetic field intensity (dB) at a point P due to current element is given by
\(d B \propto \frac{I d l \sin \theta}{r^{2}}\)
or  \(d B=\frac{\mu_{0}}{4 \pi} \frac{I d l \sin \theta}{r^{2}}\)
Thus, in vector notation,
\(d B=\frac{\mu_{0}}{4 \pi} \frac{I d 1 \times \mathbf{r}}{r^{3}}\)

\(μ_o=4π×10−7TmA−1\ or\ Wbm^{−1}A^{−1}\)

\(=2\pi R\quad or\quad R=L/2\pi ;\)
Magnetic moment=\(IA=I\pi { R }^{ 2 }\)
\(=I\pi \left( \frac { L }{ 2\pi } \right) =I{ L }^{ 2 }/4\pi \) 

Magnetic dipole moment of a current loop = niA where n = no. of turns in a current loop; i = current through the loop and A = area of each turn of the loop. Magnetic dipole moment is a vector quantity. Its direction is perpendicular to the plane of loop directed outwards for anticlockwise current in loop and is directed inwards for clockwise current in loop.

\(\overset { \rightarrow }{ F } =q\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) \)  
\(\\ or \ \left| \overset { \rightarrow }{ F } \right| =q\left| \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right| =qvBsin\theta \)
(i) F will be maximum, when \(sin\theta =1 \ or \ \theta ={ 90 }^{ o }\)  , i.e., the charged particle is moving perpendicular to the direction of magnetic field.
(ii) F will be minimum, when \(sin\theta =0 \ or \ \theta ={ 0 }^{ o } \ or \ { 180 }^{ o }\) i.e., the charged particle is moving parallel to the direction of magnetic field.

It is done by inserting laminated iron core inside the solenoid.

Right hand thumb rule states that, if we imagine a linear wire conductor to be need in the grip of the right hand such that the thumb points in the direction of current, then the curvature of the fingers around the conductor will give the direction of magnetic field lines.

The direction of \( F⃗ \) according to Right Hand rule or Fleming's Left Hand rule is along x-axis.

According to Ampere's swimming rule, magnetic field is in - z-direction.

Tesla is the SI unit of magnetic field induction or magnetic flux density at a point in the magnetic field. The magnetic field induction at a point in a magnetic field is said to be 1 tesla if one-coulomb charge while moving with a velocity of 1 m/s, perpendicular to the magnetic field experiences a force of 1 N at that point.

[M¹L¹T^-2A^-2].

The direction of magnetic field is along the +Z-axis, as per right hand rule or Fleming's Left hand rule.

Straight line.

Unit of magnetic field in cgs system is gauss and in SI is tesla or NA^-1m^-1 or weber m^-2. 1 tesla = 10⁴ gauss.

No, a stationary charge does not produce magnetic field.

The oscillator frequency should be same as proton’s cyclotron frequency.
Using Eqs. we have
B = 2π m ν / q = 6.3 x 1.67 x 10^–27 x 10⁷ / (1.6 x 10^–19) = 0.66 T
Final velocity of protons is
v = r x 2π ν = 0.6 m x 6.3 x 10⁷ = 3.78 x10⁷ m/s.
E = ½ mv ² = 1.67 x10^–27 x 14.3 x 10¹⁴ / (2 x 1.6 x 10^–13) = 7 MeV

E = 25 keV = 25 x 1.6 x 10^-16J, B = 0.2 millitesla = 0.2 x 10^-3 T,
As we know T \(=\frac{2 \pi m}{q B}\)
\( \therefore \ T=\frac{2 \pi}{\left(\frac{q}{m}\right) B}=\frac{2\left(\frac{22}{7}\right)}{\left(1.76 \times 10^{11}\right)\left(0.2 \times 10^{-3}\right)} \)
\(=1.79 \times 10^{-7} \mathrm{~s} {\left[\because \frac{q}{m}=1.76 \times 10^{11} \mathrm{Ckg}^{-1}\right]} \)

Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m
Magnitude of the magnetic field at this point is given as:
\(B=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 35}{4 \pi \times 0.2}\)
= 3.3 x 10^-5 T
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10^–5 T.

Here, I = 3A, r =  4 cm = 4 x 10^-2 m
N = 20, M = ?
\(M=NIA=NI\pi { r }^{ 2 }\) 
\(=20\times 3\times \frac { 22 }{ 7 } \left( 4\times { 10 }^{ -2 } \right) ^{ 2 }\) 
= 0.3 Am²

Here, M = 5.0 Am² , 2l = 20 cm, l = 10 cm = 0.1 m
\(m=\frac { M }{ 2l } =\frac { 5.0 }{ 0.2 } =25 \ Am\) 

Here, l = 0.5 m; N = 500; B = 2.52 x 10^-3 T
 \(B=\frac { { \mu }_{ o }NI }{ l } or \ I=\frac { Bl }{ { \mu }_{ o }N } =\frac { \left( 2.52\times { 10 }^{ -3 } \right) \times 0.5 }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times 500 }\)
\( =2.0 \ A\)

\(Given \ r=0.1m,I=5A.\)
\(\\ \therefore \ d B=\frac { { \mu }_{ 0 }I }{ 2\pi r } =\frac { 4\pi \times { 10 }^{ -7 }\times 5 }{ 2\pi \times 0.1 }\)
\(={ 10 }^{ -5 }J\)

(ii) We know, Lorentz force, F = Bqv sin\(\theta \)
Where \(\theta \) = angle between velocity of particle and magnetic field=90⁰
So,Lorentz force, F = Bqv    [sin 90⁰ = 1]
When a charged particle enters in a magnetic field in a direction normal to the field, then in this condition,
Lorentz force = Centripetal force
\(Bqv=\frac { mv^{ 2 } }{ r } \Rightarrow r=\frac { mv }{ Bq } \)

Here, I = 2.5 A, N = 100, l = 50 cm = 0.50 m
\(n=\frac{N}{l}=\frac{100}{0.50}=200\)
(i) B = \(\mu\)₀nI = 4\(\pi\)\(\times\)10^-7 \(\times\)200 \(\times\)2.5
B = 6.28 \(\times\) 10^-4 T
(ii) \(B=\frac{\mu_0 n I}{2}=\frac{4 \pi \times 10^{-7} \times 200 \times 2.5}{2}=3.14 \times 10^{-4} \mathrm{~T}\)

Here, n = 2, I = 5.0 A,
B = 0.4 x 10^-3 T, r = ?
Magnetic field at the centre of current loop is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi nI }{ r } \ or \ r=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi nI }{ B } \) 
\(\therefore \ r={ 10 }^{ -7 }\times 2\times \frac { 22 }{ 7 } \times \frac { 2\times 5.0 }{ 0.4\times { 10 }^{ -3 } } \) 
 = 157.1 x 10^-4m = 1.57 cm  

Charge on alpha particle is +2 e. The revolving alpha particle is equivalent to current loop, having current
\(I=\frac { charge }{ time } =\frac { 2e }{ t } =\frac { 2\times 1.6\times { 10 }^{ -19 }C }{ 2\quad s } \) 
= 1.6 x 10^-19 A
Magnetic field at the centre of the circular loop is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi I }{ r } =\frac { { \mu }_{ o }I }{ 2r } \) 
\(=\frac { \left( 4\pi \times { 10 }^{ -7 } \right) \times \left( 1.6\times { 10 }^{ -19 } \right) }{ 2\times 0.8 } \) 
\(=4\times \frac { 22 }{ 7 } \times \frac { { 10 }^{ -7 }\times 1.6\times { 10 }^{ -19 } }{ 2\times 0.8 } \) 
= 12.57x s10^-26 T

(i) (a): For stationary electron, \(\vec{v}=0\)
\(\therefore\) Force on the electron is \(\vec{F}_{m}=-e(\vec{v} \times \vec{B})=0\)
(ii) (d): Force on the proton \(\vec{F}_{B}=e(\vec{v} \times \vec{B})\)
Since, \(\vec{v}\) is parallel to \(\vec{B}\)
\(\therefore \quad \vec{F}_{B} \doteq 0\)
Hence proton will continue to move with velocity v along the axis of solenoid.
(iii) (b): Magnetic force on the charged particle q is 
\(\vec{F}_{m}=q(\vec{v} \times \vec{B}) \text { or } F_{m}=q v B \sin \theta\)
where \(\theta\) is the angle between \(\vec{v} \text { and } \vec{B}\)
Out of the given cases, only in case (b) it will experience the force while in other cases it will experience no force
(iv) (a) : \(\vec{F}=q(\vec{v} \times \vec{B})\)
\(=q[(2 \hat{i} \times(\hat{i}+2 \hat{j}+3 \hat{k})]=(4 q) \hat{k}-(6 q) \hat{j}\)
(v) (c): When an electric charge is moving both electric and magnetic fields are produced, whereas a static charge produces only electric field.