Current flowing in wire A, I_A = 8.0 A
Current flowing in wire B, I_B = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l = 10 cm = 0.1 m
Force exerted on length l due to the magnetic field is given as:
\(B=\frac{\mu_{0} 2 I_{A} I_{B} I}{4 \pi r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 8 \times 5 \times 0.1}{4 \pi \times 0.04}\)
= 2 x 10 ^-5 N
The magnitude of force is 2 x 10^–5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

Here, l = 0.5 m; N = 500; B = 2.52 x 10^-3 T
 \(B=\frac { { \mu }_{ o }NI }{ l } or \ I=\frac { Bl }{ { \mu }_{ o }N } =\frac { \left( 2.52\times { 10 }^{ -3 } \right) \times 0.5 }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times 500 }\)
\( =2.0 \ A\)

\(B=\frac { { \mu }_{ 0 }I }{ 2\pi r }\)
\(F=Bqv=\frac { { \mu }_{ 0 }Iqv }{ 2\pi r }\)
\( F=\frac { 4\pi { 10 }^{ -7 }\times 5\times 1.6\times { 10 }^{ -19 }\times { 10 }^{ 6 } }{ 2\pi \left( 0.1 \right) }\)
\( F=\frac { 2\times 5\times 1.6\times { 10 }^{ -20 } }{ 0.1 } \)
\(=16\times { 10 }^{ -19 }N\)

Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
 = 4 mA = 4 x 10^-3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
\(S=\frac{I_{g} G}{I-I_{g}}\)
\(=\frac{4 \times 10^{-3} \times 15}{6-4 \times 10^{-3}}\)
\(S=\frac{6 \times 10^{-2}}{6-0.004}=\frac{0.06}{5.996}\)
\(\approx 0.01 \Omega=10 \mathrm{~m} \Omega\)
Hence, a  10 mΩ shunt resistor is to be connected in parallel with the galvanometer.

(i) 36 x 10^-2 Am²
(ii) 18 x 10^-2Nm

Statement for Biot-Savart Law: The magnitude of magnetic field \(d\overrightarrow { B } \) due to current element is directly proportional to the current I, the elements length \(\left| dl \right| \) and inversely proportional to the square of the distance r of the field point. Its direction is perpendicular to the plane containing \(\overrightarrow { dl } \)and \(\overrightarrow{r}\).
\(d\overrightarrow B\alpha\frac{I\overrightarrow dl \times \overrightarrow r}{r^{ 3}}\)
Or \(d\overrightarrow B=\frac {\mu_0}{4 \pi}\frac{I\overrightarrow dl \times \overrightarrow r}{r^{ 3}}\)

The magnetic field due to \(\overrightarrow {dl}\) is given by Biot Savart law as
\(dB=\frac {\mu_0}{4\pi}.\frac { I\left| \overrightarrow { dl } \times \overrightarrow { r } \right| }{ { r }^{ 3 } } \)
Now dB_x= Db Cos \(\theta\) = \(\frac { \mu _{ 0 } }{ 4\pi } .\frac { Idl }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) } \cos { \theta } \)
\(=\frac { \mu _{ 0 } }{ 4\pi } .\frac { Idl }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) } \frac { R }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 1/2 } } \)
So, \({ B }_{ x }=\int { dB_{ s }=\frac { { \mu }_{ 0 } }{ 4\pi } } \frac { IR }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \int { dl } \)
\(=\frac { { \mu }_{ 0 }IR^{ 2 } }{ 2\left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \)
(The y-components, of the field, add up to zero,due to symmetry)
\(\therefore\)Magnetic field at P due to a circular loop
\(=B={ B }_{ x }\overrightarrow { i } =\frac { { \mu }_{ 0 }IR^{ 2 } }{ 2\left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \overrightarrow { i } \)
Explanation: A circular current loop produces magnetic field and its magnetic moment is the product of current and its area \(\overrightarrow M=\overrightarrow {LA}\)

Current sensitivity, \({ I }_{ s }=\frac { NAB }{ k } \) and
Voltage sensitivity, \(V_{ s }=\frac { NAB }{ kR } \)
Since, the resistance of the coil may vary, it implies an increase in current sensitivity may not necessarily increase voltage sensitivity.
Thus, the trajectory of both the particles will be same.

A galvanometer is said to be sensitive, if it gives a large deflection, even when a small voltage is applied cross its coil.
Current sensitivity. It is defined as the deflection produced in the galvanometer on passing unit current through its coil. Therefore,
Current sensitivity \(=\frac { \theta }{ 1 } =\frac { nBA }{ k } \) 
Voltage sensitivity. It is defined as the deflection produced in produced in the galvanometer when a unit voltage is applied across its coil. Therefore V, then Voltage sensitivity \(=\frac { \theta }{ V } \) 
If R is resistance of coil and I is current that passes through coil on applying voltage V, then \(V=IR\)
 \(\therefore \) Voltage sensitivity \(=\frac { \theta }{ IR } =\frac { nBA }{ kR } \) 
Thus, a galvanometer will be highly sensitive, if (i) n is large ; (ii) B is large ; (iii) A is large ; (iv) R is small and (v) k is small.
However, n and A cannot be increased beyond certain limit otherwise, the sixe of the galvanomert and the resistance of the instrument will become large. Therefore, B is made as large as possible. To increase B, very strong permanent magnet is used. The suspension wire is made of phosphor bronze, as for this material, k is very small. The value of k further decreases, if the wire is hammered into flat strip. In very sensitive galvanometers, quartz k, is still smaller.

The line integral of the magnetic field around some closed loop is equal to the times the algebraic sum of the currents which pass through the loop.

Since the coil is tightly wound, we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is, 
\(B=\frac{\mu_{0} N I}{2 R}=\frac{4 \pi \times 10^{-7} \times 10^{2} \times 1}{2 \times 10^{-1}}=2 \pi \times 10^{-4}=6.28 \times 10^{-4} \mathrm{~T}\) 

Using Eq. we find
r = m v / (qB) = 9 x 10^–31 kg x 3 x 10⁷ m s^–1 / ( 1.6 x 10^–19 C x 6 x 10^–4 T )
= 28 x 10^–2 m = 28 cm
ν = v / (2 \(\pi\)r) = 17 x 10⁶ s^–1 = 17 x 10⁶ Hz = 17 MHz.
E = (½ ) mv² = (½ ) 9 x 10^–31 kg x 9 x 10¹⁴ m²/s² = 40.5 x 10^–17 J
\(\approx \) 4 x 10^–16 J = 2.5 keV.

Here, n = 100, r = 8 cm = 8 \(\times\)10^-2m and I = 0.40 A
\(\therefore\) Magnetic field B at the centre,
\(\begin{aligned} B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I n}{r} & =\frac{10^{-7} \times 2 \times 3.14 \times 0.40 \times 100}{8 \times 10^{-2}} \\ \end{aligned}\)
\(\begin{aligned} =3.1 \times 10^{-4} \mathrm{~T} \end{aligned}\)

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.

Given, total number of turns, N = 500
Length of solenoid, l = 0.5 m
Current, I = 5 A
Radius, r = 1 cm = 10^-2 m
Here, \(\begin{aligned} \frac{l}{r} & =\frac{0.5}{10^{-2}}=50 \Rightarrow l>>r \\ \end{aligned}\)
\(\begin{aligned} \therefore B & =\mu_0 n I=\frac{\mu_0 N I}{l} \\ \end{aligned}\)
\(\begin{aligned} =4 \pi \times 10^{-7} \times \frac{500}{0.5} \times 5 \end{aligned}\)
= 6.28 \(\times\) 10^-3 T

(i)  Voltmeter is a high resistance device which is used to measure potential difference. Ammeter is a low resistance device which is used to measure the current in electric circuit.
(ii) Voltmeter is obtained by using a high resistance of suitable value in series of the galvanometer. Ammeter is obtained by using a low resistance shunt in parallel with a galvanometer.
(iii) Voltmeter is always connected in parallel to the conductor in circuit across which potential difference is to be determined. Ammeter is always used in series of circuit to measure the current.
(iv) The range of voltmeter can be increased or decreased. The range of ammeter can be increased but cannot be decreased.
(v) Ideal voltmeter has infinite resistance. Ideal ammeter has zero resistance.

(i) Ammeter is a low resistance galvanometer. When a low resistance shunt is connected across galvanometer, it becomes ammeter.
(ii) Voltmeter is a high resistance galvanometer. When a suitable high resistance is connected in series with the galvanometer, it becomes voltmeter.

\(=2\pi R\quad or\quad R=L/2\pi ;\)
Magnetic moment=\(IA=I\pi { R }^{ 2 }\)
\(=I\pi \left( \frac { L }{ 2\pi } \right) =I{ L }^{ 2 }/4\pi \) 

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.

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

(i) Biot-Savart law can be expressed alternatively as Ampere's circuital law.
(ii) Biot-Savart law is angle dependence law, whereas Coulomb's law in electrostatic is angle independent law.
(iii) The magnetic field due to an infinitely long current carrying wire,
\(\begin{aligned} B=\frac{\mu_0 I}{2 \pi r} \end{aligned}\)
\(\begin{aligned} \therefore \quad B & \propto I \end{aligned}\)
(iv) The magnetic field at a point on the long current carrying wire,
\(\begin{aligned} d B=\frac{\mu_0}{4 \pi} \cdot \frac{I d l \sin \theta}{r^2} \end{aligned}\)
\(\begin{aligned} =\frac{\mu_0}{4 \pi} \cdot \frac{I d l \sin 0^{\circ}}{r^2} \quad \quad \quad{\left[\because \theta=0^{\circ}\right]} \end{aligned}\)
= 0

(i) (b): As \(r_{0}=\frac{m v}{q B} \Rightarrow r^{\prime}=\frac{m\left(2 v_{0}\right)}{q B}=2 r_{0}\)
(ii) (c): As, \(T=\frac{2 \pi m}{q B}\)
Thus, it remains same as it is in dependent of velocity
(iii) (b): As \(F \perp B\)
Hence, \(a \perp B\)
\(\therefore \vec{a} \cdot \vec{B}=0\)
\(\Rightarrow \quad(x \hat{i}+2 \hat{j}) \cdot(2 \hat{i}+4 \hat{j})=0\)
\(2 x+8=0 \Rightarrow x=-4 \mathrm{~m} \mathrm{~s}^{-2}\)
(iv) (c): If the charged particle has a velocity not perpendicular to \(\vec{B},\) then component of velocity along \(\vec{B}\) remains unchanged as the motion along the \(\vec{B}\) will not be affected by \(\vec{B}\).
Then, the motion of the particle in a plane perpendicular to \(\vec{B}\) is as before circular one. Thereby, producing helical motion.
(v) (d): The force on electron \(F=q v B \sin \theta\)
As the electron is moving parallel to B
So,\(\theta=0^{\circ} \Rightarrow q v B \sin 0^{\circ}=0\)