Let the number of turns be = n
Radius = r
Length = l
Length of the wire = circumference of n turns of coil
L = n x 2\(\Pi \) r
r = L/2\(\Pi \) r
Maximum torque = nIBA = nIB \(\Pi \) r²
= nIB\(\Pi \)  (2/2\(\Pi \) n)²
= 1/n
For maximum torque n should be minimum
i.e n = 1

r = mv/qB .........(i)
Also P = mv = \(\sqrt { 2mE } \)..........(ii)
By equ (i) and equ (ii)
As the radius is \(r\frac { \sqrt { 2mE } }{ qB } \) proportional to square root of kinetic energy, so if the kinetic energy is halved the radius become \(\sqrt { 1 } /2\) times of its initial value 

(i) Let a and b be two long straight parallel conductors. I_a and I_b are the current flowing
through them and separated by a distance d. Magnetic field induction at  a point P on a conductor b due to current Ia passing through I_a is

\(B_1=\frac{\mu_02I_a}{4\pi d}\)
Now, unit length of b will experience a force as
\(F_2=B_1 I_b \times 1=B_1 I_B\)
\(\therefore F_2=\frac{\mu_0}{4\pi}\frac{2I_aI_b}{d}\)
Conductor a also experiences the same amount of force directed towards b. Hence, a and b attract each other.
(ii)

Now, let the direction of current in conductor b be reversed. The magnetic field B₂ at point P due to current I_a flowing through a will be downwards. Similarly, the magnetic field B₁ at point Q due to current I_b passing through b will also be downward as shown. The force on a will be, therefore towards the left. Also, the force on b will be towards the right. Hence, the two conductors will repel each other as shown.

Applying Ampere is circuital law for the rectangular loop abcda
\(\oint\overrightarrow B. \overrightarrow {dl}=\mu_0I\)
\(Bh=\mu_0I(nh)\)
\(B=\mu_0nI\)

As, vertical and horizontal components of magnetic fields are perpendicular to each other. so when their magnitudes are equal, resultant will divide their angle equally.
According to the question,
\(H=\sqrt { 3 } V\)
where, H and V are the horizontal and vertical components of the earth's magnetic field. If angle of dip at that place is , then
\(tan \ \delta =\frac { V }{ H } =\frac { V }{ \sqrt { 3 } V } \ [\therefore \quad H=\sqrt { 3 } V]\)
\(tan\quad \delta =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \quad \delta =\frac { \pi }{ 6 } \)
\(\therefore \) Horizontal component of the earth's magnetic field,
 \(H={ B }_{ e }\quad cos\delta \)
where,  \({ B }_{ e }\)=Earth's magnetic field
\(\frac { H }{ { B }_{ e } } =cos\delta =cos\frac { \pi }{ 6 } =\frac { \sqrt { 3 } }{ 2 }\) 
\(H:{ B }_{ e }=\sqrt { 3 } :2\)

If the current round in the face of the coil is in anti-clockwise direction, then this behaves like a North pole whereas, when it is viewed from other side, then current round in it is in clockwise direction necessarily forming South pole of magnet.

Hence, current loop have both magnetic poles and therefore, behaves like a magnetic dipole.

Here  \(B=60G,A=2\times 10^{ -4 }\quad m^{ 2 },N=40\)
\(\\ I=4mA=4\times 10^{ -3 }A,\theta =16^{ 0 }\)
\(\because\) \(\\ I=\frac { k }{ NBA } \theta =k=\frac { NBAI }{ \theta } \)
\(\\ =\frac { 40\times 60\times 2\times 10^{ -4 }\times 4\times { 10 }^{ -3 } }{ 16 }\)
\( \\ =1.2\times 10^{ -4 }\) N-m per degree

Soft iron is used to make electromagnets, because hysteresis loop for soft iron is narrow. Therefore, energy loss/volume/cycle is small.

Aluminium and sodium are paramagnetic Bismuth and copper are diamagnetic.

Ammeter is a low resistance galvanometer. The resistance of ammeter is low and that of voltmeter is high. Ammeter is connected in series and voltmeter in parallel in the circuit.

As the wire carries current I along the y-axis, so \(\overset { \rightarrow }{ l } =l\hat { j } .\) Magnitude force on wire is
\(\overset { \rightarrow }{ F } =I\left( \overset { \rightarrow }{ l } \times \overset { \rightarrow }{ B } \right) =I[l\hat { j } \times [{ B }_{ o }(\hat { i } +\hat { j } +\hat { k } )]]\) 
\(=Il{ B }_{ o }\left[ \hat { j } \times \hat { i } +\hat { j } \times \hat { j } +\hat { j } \times \hat { k } \right] \) 
\(=Il{ B }_{ o }\left[ -\hat { k } +0+\hat { i } \right] =Il{ B }_{ o }\left[ \hat { i } -\hat { k } \right] \) 
Magnitude of the magnetic force is
\(F=Il{ B }_{ o }\left[ { \left( 1 \right) }^{ 2 }+{ \left( -1 \right) }^{ 2 } \right] ^{ 1/2 }=Il{ B }_{ o }\sqrt { 2 } N\)

When a current is passed through a conductor, magnetic field is produced around the conductor. It is called magnetic effect of current. The magnetic field is in the form of concentric circular magnetic lines of force for a linear conductor carrying current. The magnetic field is in the form of parallel straight lines at the centre and concentric magnetic lines near the circular coil carrying current.

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
 \(=\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ AT \right] \left[ { LT }^{ -1 } \right] } =\left[ { ML }^{ o }{ T }^{ -2 }{ A }^{ -1 } \right] \)