Biot-Savarts law states that \(d \mathbf{B}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{I} d \mathbf{l} \times \hat{\mathbf{r}}}{|\mathbf{r}|^2}\)
Here, \(\Delta\)x = w cm
\(\because\) \(\Delta\)l = \(\Delta\)x \(\hat{\mathbf{i}}\)
\(\Rightarrow\)   I = 2 A, r = 1 m
\(\begin{aligned} \therefore \quad d B=\frac{\mu_0}{4 \pi} \cdot \frac{(2 w \hat{\mathbf{i}} \times \hat{\mathbf{j}})}{(1)^2} \\ \end{aligned}\)
\(\begin{aligned} \mathrm{I} d l=2 \times w \hat{\mathrm{i}} \\ \end{aligned}\)
\(\begin{aligned} & \because \quad \hat{\mathbf{r}}=\hat{\mathbf{j}} \Rightarrow|\mathbf{r}|=1 \mathrm{~m} \\ \end{aligned}\)
\(\begin{aligned} & \therefore \quad d \mathrm{~B}=\frac{\mu_0 w}{2 \pi} \hat{\mathbf{k}} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad|d \mathrm{~B}|=\frac{\mu_0 w}{2 \pi} \\ \end{aligned}\)
and direction along + Z-axis.

Here, I = 10 A, r = 3 cm, r = 3 x 10^-2 m
Angle subtended by coil at the centre, 
\(\theta =360^{ \circ }-90^{ \circ }=270^{ \circ }=\frac { 3\pi }{ 2 } rad\)
Magnetic field induction at O due to current through circular path ACB is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times\frac { I }{ r } \emptyset ={ 10 }^{ -7 }\times\frac { 10 }{ (3\times{ 10 }^{ -2 }) } \times\frac { 3\pi }{ 2 } \)
\(\\ B=1.57\times{ 10 }^{ -4 }T\)

A charge q projected perpendicular to the uniform magnetic field B with velocity v. The perpendicular force, F = q(v x B ), acts like a centripetal force perpendicular to the magnetic field. Then, the path followed by charge is circular as shown in the figure.

The Lorentz magnetic forces acts as centripetal force, thus
\(qvB=\frac { m{ v }^{ 2 } }{ r } \quad or\quad r=\frac { mv }{ qB }\)
here, r = radius of the circular path followed by charge projected perpendicular to the uniform magnetic field.

An ammeter measures small current and milliammeter measures small current. We know that a galvanometer can be converted into an ammeter or milliammeter using a shunt resistance S, given by
 \(S=\frac { { I }_{ g }\times G }{ I-{ I }_{ g } } \) 
It shows that the shunt resistance to be used to convert a galvanometer into a milliammeter is more than that needed to convert into an ammeter. As the shunt is connected in parallel with the galvanometer, so the effective resistance of converted galvanometer into ammeter or milliammter is , \(R=\frac { GS }{ G+S } .\) ,. It shows that milliammter will have a higher resistance than that of ammeter.

(i) A charged particle while passing through region goes undeflected (i..) without any change in velocity if \(\overset { \rightarrow }{ E } ,\overset { \rightarrow }{ v } \ and \ \overset { \rightarrow }{ B } \)  are mutually perpendicular to each other, such that the forces on charged particle due to electric field and magnetic field are equal and opposite. Due to it, they cancel out each others effect.
(ii) The force on charged particle q due to perpendicular magnetic fild,
F_m = qvB sin 90^o = qvB.
Force on charged particle due to perpendicular electric field, F_e = qE.
As the charged particle goes unde, so
qvB = qE or v = E/B elected 

The plane of coil is XY plane and field point is on the Z-axis.
\(\therefore\) Magnetic field on the axial point
\(B={{{\mu}_{0}I{R}^{2}N}\over{2({R}^{2}+{z}^{2})^{{{3}\over{2}}}}}\)
\(={{4\pi\times{10}^{-7}\times{{2}\over{\pi}}\times{(0.2)}^{2}\times100}\over{2[(0.2)^{2}+(0.2\sqrt{3})^{2}]{}^{{{3}\over{2}}}}}T\)
\(={{8\times0.04\times{10}^{-7}\times100}\over{2\times0.04\times8\times0.2}}T\)
\(=25\mu T\)

\(\because \) Horizontal component
H = B  \(cos\theta =0.4cos60^{ 0 }=0.4\times \frac { 1 }{ 2 } =0.2G\)
H = 0.2 x 10^-4T \(\left[ \because cos60^{ 0 }=1/2 \right] \)
This component is parallel to the plane of wheel. The wheel is rotating in a plane normal to the horizontal component, so it will cut the horizontal component only, vertical component of earth will contribute nothing in emf.
Thus, the emf induced is given as
\(E=\frac { 1 }{ 2 } Ht^{ 2 }\omega \)
Where, \(\omega =\frac { 2\pi N }{ t } \) and
I = length of the spoke = 50 cm = 0.5 m
\(\therefore E=\frac { 1 }{ 2 } \times 0.2\times 10^{ -4 }\times (0.5)^{ 2 }\times \frac { 2\times 314\times 120 }{ 60 } \)
E = 3.14 x 10^-5V
The value of emf induced is independent of the number of spokes as the emf's across the spokes are in parallel. So, the emf will be unaffected with the increase in spokes.

(i) Magnetic field due to straight segments is
\(B=\int { \frac { \mu _{ \circ } }{ 4\pi } } .\frac { Id1\times r }{ { r }^{ 3 } } \)

For point O, dI and r for each element of straight segments PQ and RS are parallel.
Therefore, dI x r = 0
Thus, magnetic field due to straight segments is zero.
(ii) Magnetic field at center O due to semi-circular arc
\(=\frac { Magnetic\ field\ at\ center\ of \ circular\ coil }{ 2 } \)
 \(=\frac{1}{2}\left(\frac{\mu_0 I}{2 r}\right)=\frac{\mu_0 I}{4 r}=\frac{\left(4 \pi \times 10^{-7}\right) \times 10}{4 \times 2 \times 10^{-2}}\)
[Given, I = 10 A and r = 2.0 cm = \(2\times { 10 }^{ -2 }\)m]
\(=5\pi \times { 10 }^{ -5 }T\)

\(S=\frac { { I }_{ g }G }{ I-{ I }_{ g } } =\frac { 1\times G }{ 10-1 } =\frac { G }{ 9 } \Omega \) 
Resistance of ammeter formed
\({ R }_{ p }=\frac { S\times G }{ G+S } =\frac { (G/9)\times G }{ G+G/9 } =\frac { G/9 }{ 10/9 } =\frac { G }{ 10 } \) 
\(\therefore \frac { { R }_{ p } }{ S } =\frac { G/10 }{ G/9 } =\frac { 9 }{ 10 } \)

(ii) As, \(\frac{F}{L}=\frac{\mu_0}{4\pi}.\frac{2 I_1 I_2}{r}\)
\(I_1=I_2=I A, r=1 m\)
\(\frac{F}{L}=2\times10^{-7} Nm^{-1}\)
(iii) Here, magnetic field due to the current carrying conductor at a distance d from it is
given by
\(B=\frac{\mu_0}{4\pi} \frac{2I}{d}\)
∴  Force on proton,
F = (e) (v) B sin 90°
⇒ F = evB
\(F=ev(\frac{\mu_0}{4\pi}\frac{2I}{d})\)
\(F=\frac{\mu_0}{4\pi}.\frac{2Iev}{d}\)
The proton is directed perpendicular to straight conductor and away from it.

Principle and working: A current carrying coil, placed in a uniform magnetic field, (can) experience a torque.
Consider a rectangular coil for which no. of turns = N
rea of cross-section = 1 x b = A,
Intensity of the uniform magnetic field = B,
Current through the coil = I
∴ Deflecting torque = \(BIl \times b=BIA\)
For N turns \(\tau\)  = NBIA
Restoring torque in the spring = k\(\theta\)
(k = restoring torque per unit twist)
\(\therefore NBIA=k\theta\)
\(\therefore I=(\frac{k}{NBA})\theta\)
\(\therefore I\alpha \theta\)
The deflection of the coil, is therefore, proportional to the current flowing through it.

(b) the soft iron core not only makes the field radial but also increase the strength of the magnetic field
(ii) We have
\(Current sensitivity =\frac{\theta}{I}=NBA/k\)
Voltage sensitivity =\(\frac{\theta}{V}=\frac{\theta}{IR}=(\frac{NBA}{K}).\frac{1}{R}{ 1/2}\)
It follows that an increase in current sensitivity may not necessarily increase the voltage sensitivity.

(c): When current is straight, it means the current is passing through a straight conductor, the magnetic field produced due to current through a straight conductor is in the form of concentric circular magnetic lines of force whose centres lie on the linear conductor and are in a plane perpendicular to the plane of linear conductor. It means the magnetic field is circular.