The magnetic field, due to wire I, at any point on the wire 2, is directed normal to the direction of current flow in wire 2.
Magnetic field around wire 2, due to a current I₁ in wire \({B}_{e}={{{\mu}_{0}{I}_{1}}\over{2\pi\ r}}\)
Force upon conductor carrying current due to magnetic field
\(\overrightarrow{F}=(\overrightarrow{l}\times\overrightarrow{B})\)
\(\therefore\)  Force, F₂₁ on a length 1 of wire 2 = \({I}_{2}l{{{\mu}_{0}{I}_{1}}\over{2\pi r}}\)
\(={{{\mu}_{0}{I}_{1}{I}_{2}}\over{2\pi r}}\)
Similarly,  \({F}_{12}={{{\mu}_{0}{I}_{1}{I}_{2}}\over{2\pi r}}l\)
The nature of the force is repulsive for currents in opposite direction and attractive when currents flow in the same direction.

Construction: The cyclotron is made up of two hollow semi-circular disc like metal containers, \(D_1\) and \(D_2\) called dees.
It uses crossed electric and magnetic fields.The electric field is provided by an oscillator of adjustable frequency.
Working: In a cyclotron, the frequency of the applied alternating field is adjusted to be equal to the frequency of revolution of the charged particles in the magnetic field. This ensures that the particles get accelerated every time they cross the space between the two dees. The radius of their path increases with increase in energy and they are finally made to leave the system via an exit slit.

(i) Force acting on the particle, F = Bqv
In vector form, F = q(v x B)
where, B is uniform magnetic field and v is velocity with particle which is moving.
From this equation, it is clear that direction of force is perpendicular to the plane containing both v and B. In other words, force acts perpendicular to both v and B. When velocity becomes perpendicular to force, the path of the object becomes circular.

In this case, B is assumed to act perpendicular to v. In case, B is not perpendicular to v, a component of v remains perpendicular to v. It creates circular path. The component of v parallel to B will create linear path. Here, circular path is due to v cosθ and linear path is due to v sinθ. Both when combined gives helical path.
(ii) Since, force always adjusts itself in a direction which becomes perpendicular to velocity, so only direction of velocity is changes not the magnitude. Hence, the kinetic energy of the, particle always remains constant.

(i) Ampere's circuital law states that the -line integral of magnetic field (B) around any closed path in vacuum is \(\mu_0\) times the pet current (1) threading the area enclosed by the curve. Mathematically, \(\oint B.dl=\mu_0 l\)
Ampere's law is applicable only for an Amperian loop as the Gauss's law is used for Gaussian surface in electrostatics.
(ii) According to Ampere's circuital .law, the net magnetic field is given by B = \(\mu_0\)ni
(a) The net magnetic field is given by
B_net = B₂ - B₁
= \(\mu_0 n_2 I_2 - \mu_0 n_1 I_1\)  [∵ I₂ = I₁ = I]
= \(\mu_0 I (n_2-n_1)\)
The direction is from B to A.
(b) As the magnetic field due to S₁ is confined solely inside S₁ as the solenoids are
assumed to be very long. So, there is no magnetic field outside S₁ due to current in
S₁ similarly there is no field outside S₂.
∴ B_net = 0

Moment of inertia, I₁ = I
Time period, T₁ = T
Magnetic field = B, mass of magnet = x
  
As the magnet is cut into two pieces, mass of each piece
   \(\frac { x }{ 2 } \)
Initial moment of inertia,
 \(I=\frac { 1 }{ 12 } \times mass\times { \left( length \right) }^{ 2 }\)
\(I=\frac { 1 }{ 12 } \times \ x \ \times { \left( l \right) }^{ 2 }\)
where l = length of the magnet
Moment of inertia of half piece,
\({ I }^{ ' }=\frac { 1 }{ 12 } \times \frac { x }{ 2 } { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 12 } \times \frac { x{ l }^{ 2 } }{ 4 } \)
The magnetic moment of half piece, \({ m }^{ ' }=\frac { m }{ 2 } \)
Using the formula of time period,
\(T=2\pi \sqrt { \frac { I }{ mB } } \)
\( \frac { T }{ { T }^{ ' } } =\sqrt { \frac { I{ m }^{ ' }B }{ { I }^{ ' }mB } } =\sqrt { \frac { I }{ { I }^{ ' } } .\frac { { m }^{ ' } }{ m } }\)
\( \frac { T }{ { T }^{ ' } } =\sqrt { \frac { x{ l }^{ 2 }\times 12\times 4\times 2 }{ 12\times x{ l }^{ 2 } } \times \frac { m }{ 2m } } =\sqrt { 2\times 2 }\)
\({ T }^{ ' } \ =\frac { T }{ 2 } \)
The new time period of each piece is \(\frac { T }{ 2 } \) 

\({ tan\theta }_{ 1 }=\frac { V }{ { H }_{ 1 } } ,{ tan\theta }_{ 2 }=\frac { V }{ { H }_{ 2 } } ,{ tan\theta }=\frac { V }{ { H } } \) 
As H₁ & H₂ are horizontal components in two planes at 90^o to each other,
H²​​​​​​ = H₁² + H₂²
\({ \left( Vcot\theta \right) }^{ 2 } \ ={ \left( Vcot{ \theta }_{ 1 } \right) }^{ 2 }+{ \left( Vcot{ \theta }_{ 2 } \right) }^{ 2 }\) 
\(or \ { cot }^{ 2 }\theta ={ cot }^{ 2 }{ \theta }_{ 1 }+{ cot }^{ 2 }{ \theta }_{ 2 }.\)

No. of turns, N = 850 x 5;
Area of cross section, 
\(A=\pi { r }^{ 2 }=\frac { 22 }{ 7 } { \left( 3\times { 10 }^{ -2 } \right) }^{ 2 }{ m }^{ 2 };\) 
l = 1 m; i = 5A. 
Magnetic field induction at the centre of solenoid is 
 \(B={ \mu }_{ o }Ni/l=4\pi \times { 10 }^{ -7 }\times \left( 850\times 5 \right) \times 5/1\) 
Magnetic flux = BA
\(=2.671\times { 10 }^{ -2 }\times \frac { 22 }{ 7 } \times { \left( \frac { 3 }{ 2 } \times { 10 }^{ -2 } \right) }^{ 2 }\) 
= 1.89 x 10^-5 Wb

I = 0.3 A

(ii) (a) Let the mass of proton = m; charge of proton = q, mass of a-particle = 4m
Charge of α-particle = 2q
Cyclotron frequency,
\(v=\frac{Bq}{2\pi m} \Rightarrow v \propto \frac{q}{m}\)
For proton frequency, v_p\(\propto \frac{q}{m}\)
For α-particle,
Frequency, v_a \(\propto \frac{2q}{4m}\)
or  v_a \(\propto \frac{q}{2m}\)
Thus, particles will not accelerate with same cyclotron frequency. The frequency of proton is twice than the frequency of α-particle.
(b) Velocity, \(v=\frac{Bqr}{m} \Rightarrow v \propto \frac{q}{m}\)
For proton velocity, \(v_p \propto \frac{q}{m}\)
For α-particle,
Velocity, v _a \(\propto \frac{2q}{4m}\) or v _a \(\propto \frac{q}{2m}\)
Thus, particles will not exit the dees with same velocity. The velocity of proton is twice than the velocity of α-particles.

(a) \(\overrightarrow{F}=I(\overrightarrow{l} \times \overrightarrow{B})\), where \(\overrightarrow{l}\) is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. According to Fleming's left hand rule,\(\overrightarrow{B}\) must act horizontally in a direction perpendicular to the wire carrying current.
(b) Adaptation to different situations and flexible and adjustable attitude.
(c) Avoiding unnecessary arguments in conflicting situations in everyday life.

i) Mr.Santosh is helping in nature, honest and has concern for the students to create interest in the subject.
ii) Magnetic declination, magnetic inclination and horizontal component of the earth's magnetic field.
iii) It is of the order of 10^-5 T.

Here, n = 30, G = 100 \(\Omega \) , \(\varepsilon \) = 3 V,
R = 200, K = ?
Total resistance of the circuit
= G + R = 100 + 200 = 300 \(\Omega \) 
Current in the circuit which produces full scale deflection in the galvanometer,
\({ I }_{ g }=\frac { \varepsilon }{ G+R } =\frac { 3 }{ 300 } =\frac { 1 }{ 100 } A\) 
\(K=\frac { { I }_{ g } }{ n } =\frac { 1/100 }{ 30 } =\frac { 1 }{ 3 } \times { 10 }^{ -3 }A/division\) 
\(=\frac { 1 }{ 3 } \times { 10 }^{ -3 }\times { 10 }^{ 6 }\mu \quad A/division\) 
\(=333.3\mu \ A/division\) 

Here, B = 4 x 10^-3 weber/m²;
 E = 2 x 10⁴ V/m.
As the path of moving electron is undeviated, so force on moving electron due to electric field is equal and opposite to the force on moving electron due to magnetic field, i.e., eE = evB
\(or \ v=\frac { E }{ B } =\frac { 2\times { 10 }^{ 4 } }{ 4\times { 10 }^{ -3 } } =5\times { 10 }^{ 6 } \ m/s\) 
When electron moves perpendicular to magnetic field, the radius r of circular path traced by electron is
\(or \ r=\frac { mv }{ eB } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 5\times { 10 }^{ 6 } \right) }{ \left( 1.6\times { 10 }^{ -19 } \right) \times 4\times { 10 }^{ -3 } } \) 
\(=7.11\times { 10 }^{ -3 } \ m=7.11 \ mm\) 

\({ B }_{ v }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2M\cos { \theta } }{ { r }^{ 3 } } \)
\({ \ B }_{ H }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { M\sin { \theta } }{ { r }^{ 3 } } \)
\(B=\sqrt { \begin{matrix} { { B }_{ v } }^{ 2 } & + & { { B }_{ H } }^{ 2 } \end{matrix} } =\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 4\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right] \)
\(B=\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 3\cos ^{ 2 }{ \theta } +1 \right] \)
Now B will be minimum if \(\cos { \theta } =0 \ or \ \theta ={ 90 }^{ \circ }\)
i.e. B will be minimum at magnetic equator.
(ii) Angle of dip is given by
\(\tan { \delta =\frac { { B }_{ V } }{ { B }_{ H } } } =\frac { 2\cos { \theta } }{ \sin { \theta } } =2\cot { \theta } \)
\(\delta =0,if \ \cot { \theta =0 \ or \ \theta =\frac { \pi }{ 2 } } \)
i.e. angle of dip is zero at magnetic equator(iii) \(\delta ={ 45 }^{ \circ },\cot { \theta =\frac { 1 }{ 2 } } or \ \tan { \theta =2 } \)
\( \theta =\tan ^{ -1 }{ 2is \ the \ locus. } \)

Torque = \({\overrightarrow{\mu}}_{m}\times\overrightarrow{B}\)
\(|\overrightarrow{\mu}_{m}|=nI\times A = 200 \times5\times 100 \times{10}^{-4}A-{m}^{2 }\)
= 10 A - m²
Angle between \({\overrightarrow{\mu}}_{m}\) and \(\overrightarrow{B}=90°-60°=30°\)
|Torque| = 10 x 0.2 x sin 30°
= 1 N - m

On the equator S, the vertical component of earth's magnetic field is zero.

There will be force of attraction between the straight wire and 4 cm long arm of loop nearer to the straight conductor.
\(F_1=\frac{\mu_0}{4\pi} \frac{2\times2\times1}{(2\times10^{-2})}\times(4\times10^{-2})\)  
[towards straight conductor]
\(F_1=8\times10^{-7}N\)    --- (i)
Similarly, force on other 4 cm arm of loop, away from the straight conductor
\(F_2=\frac{\mu_0}{4\pi} \frac{2\times2\times1}{(4.5\times10^{-2})}\times(4\times10^{-2})\)
\(F_2=3.55 \times 10^{-7} N\)  --- (ii)
(i) Since, F₁ and F₂ are of different magnitudes, therefore, do not form couple and hence Torque, t = 0
(ii) Net force on loop,
\(F=F_1-F_2\)
[towards straight conductor]
\(F=8\times10^{-7}-3.55\times10^{-7}\)
F =\(4.45\times10^{-7}N\)
The forces on two branches of loop are equal in magnitude and opposite in the directions, hence they balance each other.

By applying Lenz's law, we can find out direction of current in the coil. On the right hand side coil, South pole is approaching towards the coil, so at end C, South pole will be produced and on the left hand side, North pole is moving away, so at end Q coil, South pole will be produced.

(i) Paramagnetic  substance Aluminium, sodium
(Ii) Diamagnetic substance Bismuth, copper, the susceptibility of the diamagnetic materials is small and negative, i.e. -1 m < 0 whereas for paramagnetic substance the susceptibility is small and positive, -1m

Figure shows the longitudinal sectional view of long current carrying solenoid. The current comes out of the plane of paper at points marked.
The B is the magnetic field at any point inside the solenoid.
Considering the rectangular closed path abcda. Applying Ampere's circuital law over loop abcda.

\(\oint {B.dI}=\mu_0 \) ｘ(Total current passing through loop abcda)
\(\int_{a}^{b} B. dI+\int_{b}^{c} B. dI+\int_{c}^{d} B. dI+\int_{d}^{a} B. dI=\mu_0(\frac{N}{L}li)\)
where, \(\frac{N}{L}\) = number of turns per unit length
ab = cd = l = length of triangle.
\(\int_{a}^{b} B dI cos 0^{o}\int_{b}^{c} B dI cos 90^{0}+0+\int_{d}^{a} B dI cos 90{0}=\mu_0(\frac{N}{L}li)\)
\(B \int_{a}^{b} dI=\mu_0(\frac{N}{L}li) \Rightarrow Bl=\mu_0(\frac{N}{L})li\)
\(\Rightarrow B=\mu_0 (\frac{N}{L})i\)
or  \(B=\mu_0 ni\)
where, n = number of turns per unit length. This is required expression for magnetic field inside the long current carrying solenoid.

Here, I = 10 A, r = 3 cm, r = 3 x 10^-2 m
Angle subtended by coil at the centre, 
\(\theta =360^{ \circ }-90^{ \circ }=270^{ \circ }=\frac { 3\pi }{ 2 } rad\)
Magnetic field induction at O due to current through circular path ACB is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times\frac { I }{ r } \emptyset ={ 10 }^{ -7 }\times\frac { 10 }{ (3\times{ 10 }^{ -2 }) } \times\frac { 3\pi }{ 2 } \)
\(\\ B=1.57\times{ 10 }^{ -4 }T\)

Here, L = 10 cm = 0.1 m
Intensity of magnetisation, M = 10⁶ A/m
Magnetisation current, I_m = ?
Since, M = \(\frac { { I }_{ m } }{ I } \)
\(\therefore \ { I }_{ m }=M\times l\)
\(={ 10 }^{ 6 } \ A/m\times 0.1 \ m={ 10 }^{ 5 } \ A\)

As, electron is revolving clockwise, therefore conventional current due to the motion of electron will be in anti-clockwise direction.
So, according to right hand rule,magnetic field at point A will be in outward direction to the plane of circular loop.

8 x 10^-5 N

One ampere of current can be defined as the amount of current which when flows through two infinitely long parellel wires seperated by one metere produces an attractive foce 2 x 10^-7N/m.

Aluminium and calcium are paramagnetic.

No, because unlike poles of equal strength exist together.

In a galvanometer the coil having a fixed core made of non-magnetic metallic material, when oscillates, the eddy currents are set up in the metallic material which opposes the motion of the coil in the magnetic field. Due to it, the coil comes to rest at once.

As hydrogen ion is describing a circular path in a uniform magnetic field, hence force on hydrogen ion due to perpendicular magnetic field is providing the centripetal force. So
 \(Bqvsin{ 90 }^{ o }=\frac { m{ v }^{ 2 } }{ r } \ or\ r=\frac { mv }{ qB } \) 
Time period of revolution of hydrogen ion
 \(T=\frac { 2\pi r }{ v } =\frac { 2\pi }{ v } \times \frac { mv }{ qB } =\frac { 2\pi m }{ qB } \) 

The force acting on one wire due to currents through two wires is inversely proportional to the distance between them. Thus the force becomes 1/2 times if the distance between the wires is doubled.

When a point P lies on the axis of wire conductor in air, then  \(I\overrightarrow { dl } \ and \ \overrightarrow { r } \)  for each element of the straight wire conductor are parallel. Therefore, \(I\overrightarrow { dl } \times \overrightarrow { r } =0,\) so the magnetic field induction at given point is zero.

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.