Magnetic field strength, B = 6.5 x 10⁻⁴ T
Charge of the electron, e = 1.6 x 10⁻¹⁹ C
Mass of the electron, m_e = 9.1 x 10⁻³¹ kg
Velocity of the electron, v = 4.8 x 10⁶ m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = ν
Angular frequency of the electron = ω = 2πν
Velocity of the electron is related to the angular frequency as:
v = rω
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
\(e v B=\frac{m v^{2}}{r}\)
\(e B=\frac{m}{r}(r \omega)=\frac{m}{r}(2 \pi r v)\)
\(v=\frac{B e}{2 \pi m}\)
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
\(V=\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}\)
= 18.2 x 10⁶ Hz
\(\approx\) 18 MHz
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

Let \({ r }_{ 0 }=\) Maximum radius of circular path followed by charged particle (Equal to the radius of the Dees)
 \({ v }_{ 0 }=\) Maximum velocity
Since the necessary centripetal force is provided by the Lorentz magnetic force, therefore,
\( \frac { { m{ v }_{ 0 } }^{ 2 } }{ { r }_{ 0 } } =Bq{ v }_{ 0 }\) 
\({ v }_{ 0 }=\frac { Bqr_{ 0 } }{ m }\)
\( \\ \therefore \ { K.E }_{ maxi }=\frac { 1 }{ 2 } \times m\times { \left( \frac { Bqr_{ 0 } }{ m } \right) }^{ 2 }\)
\(=\frac { { b }^{ 2 }{ q }^{ 2 }{ r_{ 0 } }^{ 2 } }{ 2m } \)
This is the required result.

On the basis of their in a magnetic field, the various materials can be classified in three classes.
(i) Diamagnetic: Those materials, which when placed in a magnetic field, are feebly magnetised in a direction opposite to the magnetising field are called diamagnetic substances.A few examples of diamagnetic materials are copper, zinc, bismuth, water, sodium chloride, helium, argon etc.
When a diamagnetic substance is suspended in a magnetic field, it arranges itself in the direction of the magnetic field.
(ii) Paramagnetic: Those materials, which when placed in a magnetic field, are feebly magnetised in the direction of magnetic field, are called paramagnetic substances.A few examples of paramagnetic substances are aluminium, sodium, antimony, platinum, copper chloride, liquid oxygen etc.
When a paramagnetic substance is suspended in a magnetic field it arranges itself to the direction of magnetic field.
(iii) Ferromagnetic: Those materials which when placed in a magnetic field are strongly magnetised in the direction of the magnetising field, are ferromagnetic substances.A few examples of ferromagnetic substances are iron, nickel, cobalt, alnico, mercury etc.

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.

Time period of vibration is given by
\(T=2 \pi \sqrt{\frac{I}{M B_{H}}}\)
As the magnets are similar,
\(\therefore \quad T \propto \sqrt{\frac{I}{M}} \ \text { or } \ T \approx \frac{I}{\sqrt{m}}\)
where m is the pole strength.
Higher is the pole strength, lesser will be the time period. Thus, the magnet vibrating faster is stronger.

Given,magnetic length of bar magnet , 2l = 5 cm
l = 2.5cm
= 2.5 x 10^-2 m
Distance, d = 50 cm =0.5 m
Magnetic moment ,M = 0.40A -m²
\(B=\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2Md }{ { \left( { d }^{ 2 }-{ l }^{ 2 } \right) }^{ 2 } } \)
\(B=\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2Md }{ { d }^{ 3 } } \)
 \(= \frac { { 10 }^{ 7 }\times 2\times 0.40 }{ \left( 0.5 \right) ^{ 3 } } \)
 B = 6.4 x 10^-7 T

Magnetic field at a point on the equatorial line is given by
\({ B }_{ eq }=\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } =\frac { { 10 }^{ -7 }\times 0.40 }{ { \left( 0.5 \right) }^{ 2 } } =3.2\times { 10 }^{ -7 }T\)
Magnetic field at a point on the axial line is given by
\({ B }_{ axial }=\frac { { \mu }_{ 0 }2M }{ 4\pi { r }^{ 3 } } =\frac { { 10 }^{ -7 }\times 2\times 0.40 }{ { \left( 0.5 \right) }^{ 3 } } =6.2\times { 10 }^{ -7 }T\) 

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

When a bar magnet is cut into two equal parts, as shown in the Fig. (a), P₁ behaves as N-pole and P₂ behaves as S-pole and magnetic moment of each part of magnet becomes M/2. When pole P₂ is placed over pole P₁ as shown in Fig. (b), the net magnetic moment of the combination is zero, \(\text { i.e. } \frac{M}{2}-\frac{M}{2}=0 \)

Coulomb's law of magnetic force is inversely proportional to the squared distance between the magnetic poles and directly proportional to the product of magnetic poles.

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.

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

The working of the cyclotron is based on the fact a positively charged particle can be accelerated to a sufficiently high energy with the help of smaller values of oscillating electric field by making it to cross the same electric field time and gain with the use of strong magnetic field.

A solenoid carrying current has 
(i) maximum value of magnetic field at a point will inside the solenoid lying on the axis of solenoid.
(ii) There is a minimum value of magnetic field at a point outside the solenoid near the middle of solenoid and 
(iii) half of the maximum value at a point near the end of the solenoid.

(i) (a): It can be seen that slop of curve for wire a is greater than wire c.
(ii) (b): Inside the wire
The field of a hollow wire with constant current is homageneous
\(\text { i.e., slope } \propto \frac{I}{\pi R^{2}} \text { , }\) Current density
(iii) (c) : Wire c has the greatest radius.
(iv) (c)
(v) (a)

(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.

(a): Earth's magnetic field can be represented as the field of a huge bar magnet. If the magnet is freely suspended its north-pole points towards geographic north pole (really a south magnet pole of earth).

(c): Since \(d B \propto \sin \theta\) where \(\theta\) is angle between the direction of the flow of current and the line joining the elementary portion to the observation point which is zero in this case, so the magnetic field is also zero (because \(\sin \theta\) is equal to zero).

(c): An ammeter is a low resistance galvanometer. It is used to measure the current in amperes. To measure the current of a circuit, the ammeter is connected in series in the circuit so that the current to be measured must pass through it. Since, the resistance of ammeter is low, so its inclusion in series in the circuit does not change the resistance and hence the main current in the circuit.