Velocity selector is a device consisting of perpendicular electric and magnetic fields that can be used as a velocity filter for charged particles. It is used to measure charge to mass ratio and is also used in Mass Spectrometer.

The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v x B is along the z-axis (screw rule or right-hand thumb rule). So,
(a) for electron it will be along –z axis.
(b) for a positive charge (proton) the force is along + z axis.

\(|\mathrm{dB}|=\frac{\mu_{0}}{4 \pi} \frac{I \mathrm{~d} l \sin \theta}{r^{2}}\)
dl = Δx = 10⁻²m , I = 10 A, r = 0.5 m = y, \(\mu_{0} / 4 \pi=10^{-7} \frac{\mathrm{T} \mathrm{m}}{\mathrm{A}}\) θ = 90° ; sin θ = 1
\(|\mathrm{dB}|=\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}=4 \times 10^{-8} \mathrm{~T}\)
The direction of the field is in the +z-direction. This is so since
\(\mathrm{d} \mathbf{l} \times \mathbf{r}=\Delta x \hat{\mathbf{i}} \times y \hat{\mathbf{j}}=y \Delta x(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=y \Delta x \hat{\mathbf{k}}\)
We remind you of the following cyclic property of cross-products
\(\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} ; \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}} ; \hat{\mathbf{k}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}}\)
Note that the field is small in magnitude.

The three parameters of earth's magnetic field are : Declination, Dip and Horizontal component.

(a) No, because that would require \(\tau \) to be in the vertical direction. But \(\tau =IA\times B\), and since A of the horizontal loop is in the vertical direction, \(\tau \) would be in the plane of the loop for any B.
(b) Orientation of stable equilibrium is one where the are vector A of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.
(c) It assumes circular shape with its plane normal to the field to maximize flux, since, for a given perimeter, a circle encloses greater areas than any other shape.

Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
∴  Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, B \(=\frac{\mu_{0} 2 I}{4 \pi r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 50}{4 \pi \times 2.5}\)
= 4 x 10 ^-6 T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m
Magnitude of the magnetic field at this point is given as:
\(B=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 35}{4 \pi \times 0.2}\)
= 3.3 x 10^-5 T
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10^–5 T.

F = Il x B
F = Il B sinθ
The force per unit length is
f = F / l = I B sinθ
(a) When the current is flowing from east to west,
θ = 90°
Hence,
f = I B
= 1 x 3x 10^–5 = 3 x 10^–5 N m^–1
This is larger than the value 2 x 10^–7 Nm^–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.
(b) When the current is flowing from south to north,
θ = 0^o
f = 0
Hence there is no force on the conductor.

As the two tangent galvanometers are connected in series, they carry the same current.
 \(\therefore \ { I }_{ 1 }=\frac { 2rH }{ { \mu }_{ o }{ n }_{ 1 } } tan{ \theta }_{ 1 }={ I }_{ 2 }=\frac { 2rH }{ { \mu }_{ o }{ n }_{ 2 } } tan{ \theta }_{ 2 }\) 
\(\therefore \ =\frac { { n }_{ 1 } }{ { n }_{ 2 } } \frac { { tan\theta }_{ 1 } }{ { tan\theta }_{ 2 } } =\frac { { tan35 }^{ o } }{ { tan45 }^{ o } } =\frac { 0.7 }{ 1 } =\frac { 7 }{ 10 } \)

Magnetic field induction at P due to current through AOB and COD will be 
\({ B }_{ 1 }=\frac { { \mu }_{ o }2{ I }_{ 1 } }{ 4\pi a } { B }_{ 2 }=\frac { { \mu }_{ o }2{ I }_{ 2 } }{ 4\pi a }\)
\( Here \ \overset { \rightarrow }{ { B }_{ 1 } } and\overset { \rightarrow }{ { B }_{ 2 } } are \ perpendiculartoeachother.\quad\)
\(The \ magnitude \ of \ resultant \ magnetic \ field \ induction \ will \ be\)
\(B=\sqrt { { B }_{ 1 }^{ 2 }+{ B }_{ 2 }^{ 2 } } =\left[ \left( \frac { { \mu }_{ o }{ 2I }_{ 1 } }{ 4\pi a } \right) ^{ 2 }+\left( \frac { { \mu }_{ o }{ 2I }_{ 2 } }{ 4\pi a } \right) ^{ 2 } \right] ^{ 1/2 }\)
\(=\frac { { \mu }_{ o } }{ 2\pi a } \left( { I }_{ 1 }^{ 2 }+{ I }_{ 2 }^{ 2 } \right) ^{ 1/2 }\)

At A
Magnetic field due to conductor perpendlcuIarIy inwards \(2=B_{1}=\frac{\mu_{0}(3 I)}{2 \pi r}\)
Magnetic field due to conductor 3 perpendicular outwards \(3=B_{2}=\frac{\mu_{0}(4 I)}{2 \pi(3 r)}\)
Net magnetic field at A
\(B=B_{1}-B_{2}=\frac{\mu_{0} I}{2 \pi r}\left[3-\frac{4}{3}\right]\)
\(B=\frac{5 \mu_{0} I}{6 \pi r},\) perpendicularly inwards
(ii) Magnetic force per unit length on conductor 2 due to conductor \(1=F_{21}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 I_{1} I_{2}}{r}\)
Magnetic force per unit length on conductor 2 and due to current flowing in conductor 3 
\(F_{23}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 I_{2} I_{3}}{2 r}\)
Since forces F₂₃ and F₂₁ are in opposite directions (as shown)
∴ Net force experienced by conductor 2
\(F_{2}=F_{23}-F_{21}=\frac{\mu_{0}}{2 \pi r}\left[3 I^{2}-\frac{12 I^{2}}{2}\right]\)
\(F_{2}=\frac{3 \mu_{0} I^{2}}{2 \pi r}\) ,towards conductor 1.

For undeflected beam, \(v=\frac{E}{B}\)
i) 2 x 10⁶ m/s
(ii) F = q(E + v x B) = 1.675 x10^-5 N

Given, R₁ = 10 \(\Omega\), N₁ = 30, A₁ = 3.6 \(\times\)10^-3m²,
B₁ = 0.25 T, R₂ = 14 \(\Omega\), N₂ = 42.
A₂ = 1.8 \(\times\)10^-3 m², B₂ = 0.50 T
k₁ = k₂ (spring constants are smae)    ...(i)
(i) Using the formula of current sensitivity, \(I=\frac{N A B}{k}\)
\(\therefore \quad \frac{I_{S_2}}{I_{S_1}}=\frac{N_2 B_2 A_2 k_1}{N_1 B_1 A_1 k_2}=\frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}}\)
= 1.4     [from Eq. (i)]
(ii) Using the formula of voltage sensitivity,
\(\begin{aligned} V & =\frac{N A B}{k R} \end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{V_{S_2}}{V_{S_1}} & =\frac{N_2 B_2 A_2 k_1 R_1}{k_2 R_2 N_1 B_1 A_1} \\ \end{aligned}\)
\(\begin{aligned} =\frac{42 \times 0.50 \times 1.8 \times 10^{-3} \times 10}{14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}} \end{aligned}\)
= 1             [from Eq. (i)]

(i) (d)
(ii) (a): Using \(q v B \sin \theta=\frac{m v^{2}}{r}\)
\(r \propto \frac{1}{\sin \theta}\) for the same values of m, v, q and B
\(\therefore \frac{r_{A}}{r_{B}}=\frac{\sin 90^{\circ}}{\sin 30^{\circ}}=2 \text { or } r_{A}=2 r_{B} \text { or } r_{B}<r_{A}\)
(iii) (d): The radius of the helical path of the electron in the uniform magnetic field is
\(r=\frac{m v_{\perp}}{e B}=\frac{m v \sin \theta}{e B}=\frac{\left(2.4 \times 10^{-23} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\right) \times \sin 30^{\circ}}{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times 0.15 \mathrm{~T}}\)
\(=5 \times 10^{-4} \mathrm{~m}=0.5 \times 10^{-3} \mathrm{~m}=0.5 \mathrm{~mm}\)
(iv) (c): Here \(\vec{B}=8.35 \times 10^{-2} \hat{i} \mathrm{~T}\)
\(\vec{v}=2 \times 10^{5} \hat{i}+4 \times 10^{5} \hat{j} \mathrm{~m} / \mathrm{s}, m=1.67 \times 10^{-27} \mathrm{~kg}\)
Pitch of the helix (i.e., the linear distance moved along the magnetic field in one rotation) is given by
Pitch of the helix \(=\frac{2 \pi m v_{\|}}{q B}\)
\(=\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 2 \times 10^{5}}{1.6 \times 10^{-19} \times 8.35 \times 10^{-2}}=0.157 \mathrm{~m}\)
(v) (b): Period of revolution 
\(T=\frac{2 \pi R}{v \sin \theta} \Rightarrow T=\frac{2 \pi\left(\frac{m v \sin \theta}{q B}\right)}{v \sin \theta} \Rightarrow T=\frac{2 \pi m}{q B}\)
\(\therefore \text { Frequency, } v=\frac{1}{T}=\frac{q B}{2 \pi m}\)