(i) Magnetic field due to C at D is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ r } ={ 10 }^{ -7 }\times \frac { 2\times 3 }{ 0.10 } =6\times { 10 }^{ -6 }T\) 
(ii) Force on D, F = BI₁ lsin\(\theta \) = (6 x 10^-6) x 4 x (5 x 10^-2) x sin 90^o
 = 1.2 x 10^-6 N

Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 kV = 15 × 103 V
Electrostatic field, E = 9 × 105 V m - 1
Mass of the electron = m
Charge of the electron = e
Velocity of the electron = v
Kinetic energy of the electron = eV
\(\frac{1}{2} m v^{2}=e V\)
\(\therefore \frac{e}{m}=\frac{v^{2}}{2 V}\)            .........(1)
Since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
\(\therefore\) eV = evB
\(v=\frac{E}{B}\)                  ..............(2)
Putting equation (2) in equation (1), we get
\(\frac{e}{m}=\frac{1}{2} \frac{\left(\frac{E}{B}\right)^{2}}{V}=\frac{E^{2}}{2 V B^{2}}\)
\(=\frac{\left(9.0 \times 10^{5}\right)^{2}}{2 \times 15000 \times(0.75)^{2}}=4.8 \times 10^{7} \mathrm{C} / \mathrm{kg}\)
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He⁺⁺, Li⁺⁺, etc.

\({ B }_{ v }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2M\cos { \theta } }{ { r }^{ 3 } } \)
\({ \ B }_{ H }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { M\sin { \theta } }{ { r }^{ 3 } } \)
\(B=\sqrt { \begin{matrix} { { B }_{ v } }^{ 2 } & + & { { B }_{ H } }^{ 2 } \end{matrix} } =\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 4\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right] \)
\(B=\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 3\cos ^{ 2 }{ \theta } +1 \right] \)
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) \(\delta ={ 45 }^{ \circ },\cot { \theta =\frac { 1 }{ 2 } } or \ \tan { \theta =2 } \)
\( \theta =\tan ^{ -1 }{ 2is \ the \ locus. } \)

Magnetic field induction at a point on the axis at distance x from the centre of the circular coil carrying current is
\({ B }_{ 1 }=\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { { 2\pi nIa }^{ 2 } }{ ({ a }^{ 2 }+{ x }^{ 2 })^{ 3/2 } } \)
Magnetic field induction at the centre of the circular coil carrying current is
\({ B }_{ 2 }=\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { 2\pi nI }{ a } \)
But as per question,\({ B }_{ 1 }=\cfrac { { B }_{ 2 } }{ 8 } \)
\(\Rightarrow \ \cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { { 2\pi nIa }^{ 2 } }{ ({ a }^{ 2 }+{ x }^{ 2 })^{ 3/2 } } =\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { 2\pi nl }{ a } X\frac { 1 }{ 8 } \)
\(\Rightarrow \ \cfrac { { a }^{ 2 } }{ ({ a }^{ 2 }+{ x }^{ 2 })^{ 3/2 } } =\cfrac { 1 }{ 8a } \quad \Rightarrow { 8 }a^{ 3 }=({ a }^{ 2 }+{ x }^{ 3 })^{ 3/2 }\)
\(\Rightarrow \)  2a = (a² + x²)^1/2
\(\Rightarrow \)  4a² = a²+x²  \(\Rightarrow x=\sqrt { 3a }\)

\((i) T=2\pi r/v \ and \ i=\frac { e }{ T } =\frac { ev }{ 2\pi r } \)
\( (ii) i=ev\)
\(\\ Now \ use, \ B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi i }{ r } =\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi }{ r } ev\)

Volume of the Cubic domain
\(V={ \left( { 10 }^{ -6 } \right) }^{ 3 }={ 10 }^{ -18 }{ m }^{ 3 }={ 10 }^{ -12 }{ cm }^{ 3 }\)
Mass of cubic domain
 \(m=V\times \rho ={ 10 }^{ -12 }\times 7.9\)
\(=7.9\times { 10 }^{ -12 }g\)
Number of atoms in the domain
 \(N=\frac { 7.9\times { 10 }^{ -12 }\times 6.023\times { 10 }^{ 23 } }{ 55 } =8.65\times { 10 }^{ 10 }atoms\)
Maximum possible dipole moment will be achieved when all the domains are completely aligned.
So \({ M }_{ max }=8.65\times { 10 }^{ 10 }\times 9.27\times { 10 }^{ -24 }=8.0\times { 10 }^{ -13 }{ Am }^{ 2 }\)
Magnetisation of the domain
 \({ m }_{ max }=\frac { { M }_{ max } }{ V } =\frac { 8.0\times { 10 }^{ -13 } }{ { 10 }^{ -18 } } =8.0\times { 10 }^{ 5 }{ Am }^{ -1 }\) 

If \({ i }_{ a }\)and \({ i }_{ b }\) are the 3 antiparallel currents flowing in two parallel wires separated by a distance R, then the repulsive force experienced by unit length (1m) of either wire is
\(F=\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi R } \)
Weight per meter length of above wire
\(=0.073{ Nm }^{ -1 }\)
According to problem, the weight of lower wire is supported by upward magnetic repulsion, therefore
\(\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi R } \)\(=0.073\)
\(R=\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi \times 0.073 } \) 
\(=\frac { 4\pi \times { 10 }^{ -7 }\times 100\times 20 }{ 2\pi \times 0.073 } \)
\(=4.4\times { 10 }^{ -3 }m = 4.48mm\)

(a) Importance and production of radial magnetic field : In a radial magnetic field magnetic torque ramains maximum for all positions of the coils.
It is produced due to cylindrical pole pieces and soft iron core.
(b) Reason:
Voltmeter: This ensures that a very low current passes through the voltmeter and hence does not change (much) the original potential difference to be measured.
Ammeter: This ensures that the total resistance of the circuit does not change much and the current flowing remains (almost) at its original value.
(c) 

The coil remains suspended in radial magnetic field so that it always experiences maximum torque.
When current passes through the coil, deflection torque \(\tau(\theta)\) is produced given by
\(\tau_{deflection}=NLAB sin 90^{o}\)  --- (i)
As a result, coil rotates and phosphor bronze strip gets twisted. As a result restoring torque given by
\(\tau_{restoring}=k\theta\)   --- (ii)
where, k = torsional restoring constant
∴ In equilibrium,
\(t_{deflecting}=\tau_{restoring}\Rightarrow NIAB=k\theta\)
\(I=(\frac{k}{NAB})\theta \Rightarrow I\propto \theta\)
greater the current, greater the deflection.
(ii) In radial magnetic field, the plane of the coil is always parallel to the plane of the magnetic field and area vector of coil is perpendicular to magnetic field. It is always exerts maximum torque on the coil.
(iii) The voltmeter connected in parallel with the electrical circuit elements to measure potential difference. For exact measurement of PD voltmeter must draw minimum current which is possible only when it has high resistance. Ammeter is connected in series with the electrical circuit and current to be measured passes through it.
In order to protect the galvanometer, a feeble current must pass through the galvanometer, it is possible only when a low resistance (shunt) is connected in parallel with galvanometer to allow the major part of the current to pass through it. 

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 a charged particle moving on a circular path of radius R in a uniform magnetic field, the Lorentz magnetic force F (= qvB) acting on the particle, provides the required centripetal force for its circular motion. It means the Lerentz force acts along the radius towards the centre of circular path. While moving on a circular path, the small displacement \(\overset { \rightarrow }{ dr } \) of the charged particle is always perpendicular to Lerentz force, i.e., \(\theta ={ 90 }^{ o }\) , therefore work done
\(dW=\overset { \rightarrow }{ F } .\overset { \rightarrow }{ dr } =F\quad dr\quad cos{ 90 }^{ o }=0.\) 
Since the velocity of the charged particle, moving on a circular path is acting tangentially to the path whose direction is changing continuously in circular motion of the particle, therefore the momentum of the particle is changing.

No, because the source of magnetic field is a magnetic dipole, but the source of electric field is an electric charge.

When a straight wire is bent into the form of a circular coil of N turns, the length of the wire is equal to circumference of the coil multiplied by the number of turns. Let the radius of coil be r.

As, the wire is bent round in the form of a coil having N turns.
\(\therefore\) N \(\times\) Circumference of the coil = Length of the wire
\(\Rightarrow\)  N \(\times\)(2\(\pi\)r) = L
\(\Rightarrow \ r=\frac{L}{2 \pi N}\) ....(i)
Magnetic field at the centre due to N turns of a coil is given by
\(\begin{aligned} B & =\frac{\mu_0(N \mathrm{I})}{2 r}=\frac{\mu_0(N \mathrm{I})}{2\left(\frac{L}{2 \pi N}\right)} \\ \end{aligned}\)     [from Eq. (i)]
\(\begin{aligned} =\frac{\mu_0 \pi N^2 \mathrm{I}}{L} \end{aligned}\)
The direction of magnetic field is perpendicular to the plane of loop and entering into it.

\(\overset { \rightarrow }{ F } =q\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) \)  
\(\\ or \ \left| \overset { \rightarrow }{ F } \right| =q\left| \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right| =qvBsin\theta \)
(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.

Magnetic field induction at the centre of the circular coil carrying current is 
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi nI }{ r } \ i.e.\ B\propto \frac { n }{ r } \)
\( \therefore \frac { B' }{ B } =\frac { 2n }{ \left( r/2 \right) } \times \frac { r }{ n } =4 \ or \ B'=4B\)

[M¹L¹T^-2A^-2].