\(Here,B=0.3 \ T,\theta ={ 30 }^{ o },\tau =0.06N-m,M=?\) 
\(As \ \tau =MB\quad sin\theta ,\ M=\frac { \tau }{ Bsin\theta } =\frac { 0.06 }{ 0.3sin{ 30 }^{ o } } \) 
= 0.4 Am²

Here, pole strength, \(m=\frac { M }{ l } \) 
If r is radius of semicircle, then
 \(\pi r = l, \ r=\frac { l }{ \pi } \) 
New magnetic moment M' = m X 2r
 \(=\frac { M }{ l } .\frac { 2l }{ \pi } =\frac { 2M }{ \pi } \) 

F = qv B sin\(\theta \) = (1.6 x 10^-19) x (3 x 10⁷) x (10) x sin90^o = 4.8x10^-11 N. Here \(\overset { \rightarrow }{ B } \) is directed east wards; current is southwards which is opposite to the motion of electrons. Now apply Fleeming's Left hand rule, the direction of \(\overset { \rightarrow }{ F } \) is vertically upwards.

By using R = 2 \(\times\) 10⁴\(\Omega\) in series

Iron, cobalt, nickel, steel etc. are magnetic substances. Brass, paper, wood etc. are non-magnetic substances.

The shunt is a low resistance connected in parallel to a galvanometer to protect it from the strong currents in the circuit. The galvanometer shows full scale deflection with a very small current. If the whole of the strong current in an electric circuit passes through galvanometer, it may get damaged. The shunt allows only a small part of the current to flow through the galvanometer by pass to the major part of the current.

Torque on the coil,
\(\tau =IAB\quad b \ cos\theta =IlbB \ cos\theta \)
(i)Torque is minimum if minimum, i.e.,
i.e., the plane of coil is set perpendicular to the direction of magnetic field.
(ii) Torque is maximum if \(cos\theta =1 \ or \ \theta ={ 0 }^{ o }\) It means the plane of coil parallel to the direction of magnetic field.

When an electron is accelerated in a cyclotron, very soon it acquires a very high velocity. Due to it, its mass increases with velocity according to relation \(m={ m }_{ o }/\sqrt { \left( 1-{ v }^{ 2 }/{ c }^{ 2 } \right) } \) . Therefore the time taken by electron to describe semicircular path inside the dee of a cyclotron, \(t=\pi m/Bq,\) also increases with the increase of m. Due to it, the electron does not arrive in the gap between the two dees exactly at the instant, the polarity of the two dees is reversed. As a result of it, the electron goes out of steps with the oscillating electric field and hence can not be accelerated by cyclotron.

Lorentz magnetic force on a moving charged particle in the magnetic field is 
\(\overset { \rightarrow }{ F } =q\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) \) 
This force is acting perpendicular to the plane containing \(\overset { \rightarrow }{ v } \) and \(\overset { \rightarrow }{ B } \) , and is directed as given by Right Hand rule. Since work done,
\(\overset { \rightarrow }{ W } =\overset { \rightarrow }{ F } .\overset { \rightarrow }{ s } =Fscos\theta ,\) 
where   is displacement of charged particle in magnetic field. Here, angle \(\theta ={ 90 }^{ o }\) , so 
W = Fs cos 90^o
= 0.

A moving coil galvanometer cannot be used to detect a.c. in a circuit, since it measures the average value of current and the average value of a.c. over a complete cycle is zero.

The resistance of milliammeter is greater than that of ammeter. The resistance of voltmeter is greater than that of millivoltmeter.

It concentrates the magnetic field and helps in making the magnetic field radial.

The principle of galvanometer is based on the interaction of current and magnetic field.

Wb/m² is the SI unit of magnetic field induction B, which is a vector quantity.

\(Kinetic \ energy, \ K=\frac { 1 }{ 2 } { mv }^{ 2 } \ or \ v=\sqrt { \frac { 2K }{ m } }\)
\(Radius \ of \ curvature,\)
\(\\ r=\frac { mv }{ qB } =\frac { m }{ qB } \sqrt { \frac { 2K }{ m } } =\frac { \sqrt { 2mK } }{ qB } \)
\( Therefore,\ r\propto \sqrt { k } ,\)
\( So \ \frac { r' }{ r } =\sqrt { \frac { K/2 }{ K } } =\frac { 1 }{ \sqrt { 2 } } or \ r'=\frac { r }{ \sqrt { 2 } }\)

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.

Magnetic flux density at a point in a magnetic field means magnetic field induction at that point. It is defined as the force experienced by a unit charge while moving with a unit velocity, perpendicular to the direction of magnetic field at that point. Force experienced by the charged particle having charge q moving with velocity \(\overset { \rightarrow }{ v } \) through a magnetic field \(\overset { \rightarrow }{ B } \) is given by
\( \left| \overset { \rightarrow }{ F } \right| =q\left| \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right| =qvBsin\theta\)  
\(or \ B=\frac { F }{ qvsin\theta } \)
The SI unit of B is tesla, where 1 tesla is the magnetic flux density at a point if 1 coulomb charge while moving with a velocity of 1 ms^-1, perpendicular to a magnetic field experiences a force of 1 N at that point.
The dimensional formula of B
 \(=\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ AT \right] \left[ { LT }^{ -1 } \right] } =\left[ { ML }^{ o }{ T }^{ -2 }{ A }^{ -1 } \right] \)

A solenoid consists of an insulating long wire closely wound in the form of helix. Its length is very toroid is a hollow circular ring on which a large number of insulated turns of a metallic wire are closely wound. Infact, toroid is an endless solenoid.

A current loop behaves as a bar magnet because
 (i) one face of current loop behaves as a south pole and the other face as north pole.
(ii)  it possesses a magnetic dipole moment (M = IA) and 
(iii) it experiences a torque in an external magnetic field, which tends to align the axis of the loop along the direction of magnetic field as bar magnet does.

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.

Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection,  I_g = 3 mA = 3 x 10^-3 A
Range of the voltmeter is 0, which needs to be converted to 18 V.
therefore, V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
\(R=\frac{V}{I_{\mathrm{g}}}-\mathrm{G}\)
\(=\frac{18}{3 \times 10^{-3}}-12=6000-12=5988 \Omega\)
Hence, a resistor of resistance 5998 Ω is to be connected in series with the galvanometer.

\(\oint \overrightarrow B.\overrightarrow{dl}=\mu_0\Sigma i\)
\(\int _{ a }^{ b }{ \overrightarrow { B } .\overrightarrow { dl } + } \int _{ b }^{ c }{ \overrightarrow { B } .\overrightarrow { dl } + } \int _{ c }^{ d }{ \overrightarrow { B } .\overrightarrow { dl } + } \int _{ d }^{ a }{ \overrightarrow { B } .\overrightarrow { dl } =\mu _{ 0 }I(nh) } \)
\(Bh+0+0+0=\mu_0I(nh)\)
\(B=\mu_0nI\)
(b) (Anyon difference ~ In a toroid, magnetic lines do not exist outside this body.
￫Toroid is close whereas the solenoid is open on both sides
￫ Magnetic field is uniform inside a toroid whereas for solenoid, it is different at the two ends and the centre.

Strengthing of magnetic field: (Anyone) 1
(i) By inserting a ferromagnetic substance inside the solenoid
(ii) By increasing the amount of current through the solenoid

(i) The shunt is a low resistance connected in parallel with the galvanometer. Therefore, the combined resistance is less than that of the shunt.
(ii) Let I be the total current passing through the shunt galvanometer. Let \({ I }_{ g }\),\({ I }_{ s }\) be the currents through galvanometer of resistances G and shunt of resistance S respectively. As current divides itself in the inverse ratio of the resistances, therefore, fraction of current passing through galvanometer
= \(\frac { { I }_{ g } }{ I } =\frac { S }{ G+S } =\frac { 3 }{ 30+3 } =\frac { 1 }{ 11 }\)
\(Fraction \ of \ current \ passing \ through \ shunt=\frac { { I }_{ s } }{ I } =\frac { G }{ G+S } =\frac { 30 }{ 30+3 } =\frac { 10 }{ 11 } \\ \)
(iii) From the above study, we find that the current always divides itself in the inverse ratio of resistances. The major part of current flows through the shunt, which is the path of low resistance. Similarly in life, if different paths are available to reach a destination, then one selects a path of least resistance to reach there. This would save both, time and energy.

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.