Force acting on changed particle is zero, as v of changed particle remains constant  
$$F= q(v\times B) $$

Electric current is basically flow of +ve charge (assumed) and force on a charge moving in any direction with velocity $$\vartheta$$ is given by

Since the point P lies on the axis of straight part OA and hence magnetic field due to this part will be zero.

Magnetic field at the centre of a circular coil
$$B_{Circular} = \dfrac {\mu_{0}}{4\pi} \dfrac {2\pi i}{r} = \dfrac {\mu_{0}}{4\pi} \dfrac {4\pi^{2}i}{L} \left [\because L = 2\pi r\ and\ r = \dfrac {1}{2\pi}\right ]$$
Magnetic field at the centre of square coil
$$B_{Square} = \dfrac {\mu_{0}}{4\pi}\dfrac {8\sqrt {2}i}{a}= \dfrac {\mu_{0}}{4\pi} \dfrac {32\sqrt {2}i}{L}\left [\because L = 4a\ and\ a = \dfrac {L}{4}\right ]$$
Hence, $$\dfrac {B_{Circular}}{B_{Square}} = \dfrac {\dfrac {mu_{0}}{4\pi}\dfrac {4\pi^{2}i}{L}}{\dfrac {\mu_{0}}{4\pi}\dfrac {32\sqrt {2}i}{L}} = \dfrac {\pi^{2}}{8\sqrt {2}}$$.

Magnetic field due to straight wire above O is zero
i.e., $$B_1=0$$(since, $$\theta =0^o$$)
The magnetic field due to semi circular part
$$B_2=\left(\displaystyle\frac{\mu_0nI}{2r}\right)=\frac{\mu_0(1/2)I}{2r}=\frac{\mu_0I}{4r}$$
The magnetic field due to lower straight portion
$$B_3=\displaystyle\frac{1}{2}\frac{\mu_0I}{2\pi r}$$(upward)
Net magnetic field $$B=B_1+B_2+B_3$$
$$=0+\displaystyle\frac{\mu_0I}{4r}+\frac{\mu_0I}{4\pi r}$$(upwards).

Here, $$\theta = 60^0, B_1 = 1.2 \times 10^{-2}T$$

The magnetic material can be classified as 

$$\overset{\rightarrow}{F} = \overset{\rightarrow}q ( \overset{\rightarrow}{v} \times \overset{\rightarrow}{B}) \,\, \,\, \therefore F = qv B sin \theta$$
which shows magnetic force is velocity dependent due to which it differs from one inertial frame to another.

Here, $$\theta = 30^0, B = 0.35T$$
$$T = 4.5 \times 10^{-8}Nm$$

It is given that

Number of turns $$n=4$$

Length of the uniform conducting wire $$l=10\,a$$

Area of an equilateral triangle $$A=\dfrac{\sqrt{3}}{4}{{a}^{2}}$$ where a is the side of triangle

Magnetic moment

$$ M=nIA $$

$$ M=4\times I\times \dfrac{\sqrt{3}}{4}{{(10)}^{2}} $$

$$ M=\sqrt{3}{{a}^{2}}I $$

Here, $$M=0.355$$J $$T^{-1}$$
$$B=0.5\times 10^{-2}$$T, v$$=2$$Hz
As $$v=\dfrac{1}{2\pi}\sqrt{\dfrac{MB}{I}}$$
$$\therefore v^2=\dfrac{1}{4\pi^2}\dfrac{MB}{I}$$
$$\Rightarrow I=\dfrac{MB}{4\pi^2v^2}=\dfrac{0.355\times 5\times 10^{-2}}{4\times (3.14)^2\times 2^2}=\dfrac{1775\times 10^{-2}}{157.75}$$
$$=1.13\times 10^{-4}$$kg $$m^2$$

Ampere 's circuital law states that relationship between thecurrent and the magnetic field

$$\displaystyle E = \frac{1}{2} mv^2 \Rightarrow v = \sqrt{ \dfrac{2E}{m} }$$
$$\displaystyle r = \frac{mv}{Be} = \frac{m}{Be} \sqrt{ \dfrac{2E}{m} } =  \dfrac{\sqrt{2mE}}{Be}$$
$$\displaystyle r = \frac{\sqrt{2 \times 1800 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31}}}{1.6 \times 10^{-19} \times 0.4} = 3.58 \times 10^{-4} m$$

On the axis of a magnetic dipoles, magnetic induction is
$$B=\dfrac{\mu_o}{4\pi}\times \dfrac{2m}{r^3}$$

The magnetic moment is given by
$$| \bar m| = NIA = NI \pi r^2$$

According to Biot-Savart's law, the magnetic field
$$\displaystyle \overset{\rightarrow}{B} = \frac{\mu_o}{4 \pi} . \frac{q (\overset{\rightarrow}{v} \times \overset{\rightarrow}{r} ) }{r^3}$$
The direction of $$\overset{\rightarrow}{B}$$ will be along $$\overset{\rightarrow}{v} \times \overset{\rightarrow}{r}$$ i.e. perpendicular to the plane containing $$\overset{\rightarrow}{v}$$ and $$\overset{\rightarrow}{r}$$.

Torque, $$\bar{\tau}=\vec{m}\times \vec{B}=mB\sin\theta$$
and magnetic potential energy $$U_m=\vec{m}\cdot \vec{B}=mB\cos \theta$$
at $$\theta =0^o$$ the dipole will be in most stable position
$$\tau =mB\sin\theta =mB\sin 0^o=0$$
and $$U_m=-mB\cos\theta =-mB\cos \theta^o=-mB$$

In this situation if the particle is thrown in x-y plane (as shown in figure) at some angle θ with velocity v, then we have to resolve the velocity of the particle in rectangular components, such that one component is along the field (v cosθ) and other one is perpendicular to the field (v sinθ). We find that the particle moves with constant velocity v cosθ along the field. The distance covered by the particle along the magnetic field is called pitch.

The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by

$$p = T ( v \cos \theta ) = 2 p \dfrac { m } { q B } ( v \cos \theta )$$

For given pitch $$p$$ correspond to charge particle, we have

$$\quad \dfrac { q } { m } = \dfrac { 2 \pi v \cos \theta } { q B } =$$ constant

Here in this case, charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B, LHS for two particles should be same and of opposite sign.

Therefore, $$\left( \dfrac { e } { m } \right) _ { 1 } + \left( \dfrac { e } { m } \right) _ { 2 } = 0$$

$$\dfrac{\mu_op}{4\pi x^2}=\dfrac{\mu_0p}{4\pi \times 1^2}+\dfrac{\mu_0\times p}{4\pi \times 2^2}+\dfrac{\mu_0p}{4\pi\times 4^2}+$$________

Let r be the radius of the circular loop.
$$\therefore$$ A = $$\displaystyle \pi r^2$$   or $$r$$ = $$\displaystyle \sqrt{\frac{A}{\pi}}$$
Magnetic field at the centre of the loop is 
$$B$$ = $$\dfrac{\mu_0I}{2r}=\dfrac{\mu_0I}{2\sqrt{\dfrac{A}{\pi}}}$$    or    $$I=\dfrac{2B}{\mu_0}\sqrt{\dfrac{A}{\pi}}$$
The magnetic moment of the loop is
M= $$IA$$ = $$\dfrac{2B}{\mu_0}\sqrt{\dfrac{A}{\pi}}A$$ = $$\dfrac{2BA\sqrt A}{\mu_0\sqrt\pi}$$

Here, $$B=900$$ Gauss $$=900\times 10^{-4}$$T.
$$=9\times 10^{-2}$$T
$$\tau =0.02$$N m and $$\theta =30^o$$
$$\because \tau =mB\sin \theta$$
$$\Rightarrow 0.02=m\times 9\times 10^{-2}\times \sin 30^o$$
$$0.02=9\times 10^{-2}\times \dfrac{1}{2}\times m$$
$$m=\dfrac{0.02\times 2}{9\times 10^{-2}}=0.44$$A $$m^2$$.

Length $$l$$, long wire is twisted in the state of half circle of span $$r$$, at that point, length of the wire is calculated as follows:

Given

Given,

Magnetic field produced by charges moving with velocity $$\bar v$$, at a distance r is $$ \bar B$$ = $$\left ( \dfrac{\mu_0}{4\pi } \right )$$.q$$\dfrac{\bar v \times \hat r}{r^2}$$

For each of the 12 wires, perpendicular distance, $$\alpha$$ and $$\beta$$ are the same.
B=$$ \dfrac{\mu_0 i}{4\pi \dfrac{l\sqrt3}{2}}[sin\alpha + sin\beta]$$
$$\alpha =- 30^0$$
$$\beta =+60^0$$
B=$$\dfrac{\mu_0i}{2\pi l \sqrt3}\Big[ \dfrac{\sqrt3}{2}- \dfrac{1}{2}\Big]\times 12$$
$$=\dfrac{\mu_0 i}{\pi l} ( 3- \sqrt3)$$

We know,

After bending, distance between poles= $$2r$$
Earlier distance =$$ l(\pi r)$$
$$\dfrac{M'}{M} =\dfrac{m\times 2r}{m\times \pi r}=\dfrac{2}{\pi}$$
$$\therefore M'= M\times \dfrac{2}{\pi}$$

Movement of charge in a straight line results in a magnetic field being generated from the relation $$F=qvB\sin\theta$$

Polarity induced $$X$$ is shown in fig.

$$R= \dfrac{mv}{qB} =\dfrac{m}{qB}\sqrt{\dfrac{2(KE)}{m}}= \dfrac{\sqrt{2m}}{qB}\sqrt{KE}$$

$$V=\dfrac l{t}=1m/s$$

$$\vec{DA}=-2\cos 30\hat i-2\sin 30\hat k=(-\sqrt{3}\hat i-\hat k)$$

$$\quad In\quad this\quad the\quad magnetic\quad field\quad due\quad to\quad each\quad straight\quad conductor\quad is\\ \quad \quad \quad being\quad added\quad and\quad is\quad in\quad the\quad inward\quad direction\\ \quad \quad the\quad magnetic\quad field\quad due\quad to\quad one\quad straight\quad conductor\quad B=\quad \frac { { \mu  }_{ 0 }i(sin\alpha \quad +sin\beta ) }{ 4\pi r } \\ \quad \quad Here\quad both\quad \alpha =\beta ={ 72/2 }^{ 0 }\quad degrees\\ \quad \quad \quad \quad \quad Now\quad putting\quad value\quad B=\quad \frac { { \mu  }_{ 0 }i(2sin{ 72/2 }^{ 0 }) }{ 4\pi r } \\ \quad \quad \quad \quad \quad \quad Now\quad total\quad magnetic\quad field\quad at\quad the\quad centre\quad is\quad 4B=\quad 4\frac { { \mu  }_{ 0 }i(2sin{ 72/2 }^{ 0 }) }{ 4\pi r } \\ \quad \quad \quad \quad \quad \quad where\quad r\quad is\quad \frac { 4a }{ 5 } \quad therefore\quad total\quad magnetic\quad field\quad becomes=\frac { 5{ \mu  }_{ 0 }i(2sin{ 72/2 }^{ 0 }) }{ 4\pi a } \\ \quad therefore\quad option\quad A$$

Biot-savart law is given by

We use a standard right handed coordinate system $$\left( {r,\phi ,z} \right)$$

When turned into coil, its radius becomes;

When a charged particle enters a magnetic field $$B$$ its kinetic energy remains constant as the force exerted on the particle is:
$$F=q\overrightarrow { V } \times \overrightarrow { B } $$
is perpendicular to $$\overrightarrow {V}$$, so the work done by $$\overrightarrow {B}=0$$. This does not cause any change in kinetic energy.

$$\textbf{Hint:}$$ Use the magnetic dipole moment formula.

Given,

Charge density of each rod, $$\lambda $$ (charge per unit length)

$$K=\dfrac{1}{4\pi {{\varepsilon }_{o}}}$$

Separation between rod, $$d$$

Electric field (at distance d) by a uniform charged rod, $$E=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2\lambda }{d}=\dfrac{2K\lambda }{d}$$

Total charge on second rod $$\text{=}\ \text{charge}\ \text{density}\ \times\ \text{length}\ \text{=}\ \lambda L$$

Force on charged rod placed in electric field, $$F=(\lambda L)E=\dfrac{2KL{{\lambda }^{2}}}{d}$$

Force per unit length due to rod is  $$\dfrac{F}{L}=\dfrac{2K{{\lambda }^{2}}}{d}$$ 

In cyclotron output energy of particles coming out for deutron is

$$ {{E}_{d}}=\dfrac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2{{m}_{D}}} $$

$$ {{E}_{P}}=\dfrac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2{{m}_{P}}} $$

Taking ratios,

$$ \dfrac{{{E}_{d}}}{{{E}_{p}}}=\dfrac{{{m}_{p}}}{{{m}_{d}}} $$ $$ \because {{m}_{p}}=1.6\times {{10}^{-27}}\,\,\And \,\,{{m}_{d}}=3.2\times {{10}^{-27}} $$

$$ \dfrac{20}{{{E}_{P}}}=\dfrac{1.6\times {{10}^{-27}}}{3.2\times {{10}^{-27}}} $$

$$ {{E}_{p}}=40\,\,MeV $$

 

after passing through a magnetic field, the magnitude of its mass and velocity of the particle remains same, so its energy does not change that is kinetic energy will remain T.

No. of oscillation per min=$${1 \over {2\pi }}\sqrt {{{M{B_H}} \over I}} $$

$$n\alpha \sqrt {M{B_H}} $$

$$M \to 4times$$

$${B_H} = 2times$$

$$v \to \sqrt 8 times$$

$${v^1} = \sqrt 8 V$$

$$ = 2\sqrt 2 n$$

The formula is $$B = \dfrac{\mu _o}{4\pi} I \dfrac{d\vec l \times \vec r}{r^3}$$