Moving Charges And Magnetism - Class 12 Engineering Physics - Extra Questions

A charged particle experiences a force when moving through a magnetic field.

It the current is increased in the conductor the deflection of the compass needle increases. This is because the strength of magnetic field varies directly as the magnitude of the electric current, or the current passing through the wire.

Radius of circle path in transverse magnetic field
$$r=\dfrac{mv}{qB} \propto \dfrac{m}{q}$$ for same $$v$$ and $$B$$
For $$\alpha-$$ particle $$\left(\dfrac{m}{q}\right)_{\alpha}=\dfrac{4m_p}{2e}=\dfrac{2m_p}{e}$$ where $$m_p$$ is mass of proton.
For $$\beta-$$ particle $$\left(\dfrac{m}{q}\right)_{\beta}=\dfrac{\dfrac{1}{1840}m_p}{e} = \dfrac{1}{1840} \left(\dfrac{m_p}{e} \right)$$
Clearly $$\beta-$$ particle has smallest value of $$\dfrac{m}{q};$$ so $$\beta-$$ particle will describe the smallest circle.

Radial Magnetic Field: It means that the plane of the coil is always parallel to the lines of forces so in every position $$\theta =90^o$$.

Underlying principle of moving coil galvanometer is that,when a current carrying conductor is placed inside a magnetic field it experiences magnetic force.

When a charged particle enters in a magnetic field at an angle other than $$90^{o}$$  then a component of velocity takes it under linear motion and other component , under circular motion ,and the combined effect is a helical motion .

Galvanometer works on the principle of tangent law in magnetism. 

The nature of magnetic field in a moving coil galvanometer is radial.

According to the Right hand thumb rule, the direction of the magnetic field is perpendicular to the direct of the current and if a magnetic compass is brought closer to the current carrying conductor, the deflection is maximum. But when the needle is placed near the point A on the plane as that of the paper as given in the figure, there is no deflection. Hence, the current carrying conductor needs to be placed in the same plane as that of the paper, for the magnetic compass to attain no deflection.

emf = rate of change of flux = $$ \dfrac{ d (\pi r^2 B) }{dt} = \pi r^2 \dfrac{dB}{dt} = \pi r^2 K $$ 
current induced = $$ \dfrac{emf}{R} = \dfrac{  \pi r^2 K}{R} $$
total charge that will flow = $$ \int_0^{C/K}{i dt} =\int_0^{C/K}{\dfrac{  \pi r^2 K}{R} dt}= \dfrac{  \pi r^2 K}{R} \dfrac{C}{K}    $$coulombs

Current in wire A, $$I_1=8\ A$$

The force between two wires is given by the relation,

The direction of force experienced by the wire 'a' is toward the wire 'b' as shown.
Similarly the direction of force experienced by the wire 'b' is toward the wire Thus, the force is attractive.
If two long wires are placed in vacuum at a separation of 1 m, one Ampere would be defined as the current in each wire that would produce a force of per unit length of wire.

Definition of $$1\ ampere$$ :- 

The force per unit length is given by:

Consider that magnetic field density at point P is zero due to current carrying conductors.
According to right hand palm rule No. 1, direction of magnetic field will be perpendicular to the plane of the paper
(1) Downward due to current flowing in conductor X.
(2) Upward due to current flowing in conductor Y.
Since at point $$P, B=0$$
$$B_1=B_2$$
$$\dfrac{{\mu}_0}{2\pi}\dfrac{L_1}{x}=\dfrac{{\mu}_0}{2\pi}\dfrac{L_2}{24-x}$$
$$\dfrac{5}{x}=\dfrac{7}{24-x}$$
$$5(24-x)=7x$$
$$5\times 24 = 12x \Rightarrow x = 10 cm$$ from conductor X.

Induced emf$$=e = B\ vl = 0.1\times 5\times 0.4 = 0.2\ volt$$

The Ampere is that current which when flowing in two infinite parallel wires one metre apart produces a force between them of 2 x 10-7 N/m.

(i) Ampere : One ampere is the current, if it is maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one meter apart in vacuum, would produce between these conductors a force equal to $$2\times 10^{-7}$$ Newton per metre of length.

(ii) Ideal transformer : A transformer which does not have any loss in it, means no core losses, copper losses and any other losses, is called ideal transformer. For ideal transformer $$P_{out}=P_{in}$$, so efficiency $$\eta = 100\%$$ (But practically, efficiency of transformer lies between $$70\%-90\%$$) Ideal transformer is a theoretical concept (imaginary transformer), practically it is not possible.

When the current through the conductor Y is reversed, at $$Q, B=B_1+B_2$$ (So in between neutral point is not possible)
at P,
$$B_P=B_1(\uparrow)-B_2(\downarrow)$$
$$O=\dfrac{{\mu}_0}{2\pi}\times \dfrac{5}{r}-\dfrac{{\mu}_0}{2\pi}\times \dfrac{15}{(24+r)}$$
$$\dfrac{5}{r}=\dfrac{15}{24+r}$$
$$3r=24+r$$
$$r=12 cm$$ i.e. to the left of X
at point R
$$B=B_1(\downarrow)-B_2(\uparrow)$$
$$O=\dfrac{{\mu}_0}{2\pi}\dfrac{5}{(r+24)}-\dfrac{{\mu}_0}{2\pi}\dfrac{15}{r}$$
$$\dfrac{5}{r+24}=\dfrac{15}{r}$$
$$3r+72 = r$$
$$r=-36 cm$$ which is not possible.
Hence neutral point will lie at a distance of $$r=12 cm$$ to the left of conductor X.

Place a current carrying insulated wire in between the poles of a horse-shoe magnet. As the current passes it creates a magnetic field around it given by Flemming’s Right Hand Rule. Because the magnetic field created by the electric current in the wire is changing directions around the wire, it will repel both poles of the magnet by bending away from the wire. Depending on which pole is up, the wire will bend away from the magnet or farther into the “U”.

The magnetic field due to one current carrying wire on the other wire separated by a distance $$d$$ is given as 

Magnetic effect of electric current - " The production of magnetic field around a conducting wire when electric current is allowed to flow through it, is called magnetic field of electric current."

When product is maintained a negaive value, Force will be repulsive (+ve)

$$(i)$$ Let $$\vec { B } ={ B }_{ 0 }(-\hat { K } )$$

Let charge on particle is $$q$$ and mass of particle is $$m$$. Moving with velocity $$v$$, perpendicular to magnetic field $$(B)$$ now. Radius of circle related as $$\cfrac { mV^{ 2 } }{ r } =qVB$$

Since current in both the wire is flowing in same direction, so both wire will attract each other.
Force per unit length acting on the wire B due to A    $$F = \dfrac{\mu_o I_1I_2 }{2\pi r}$$
where  $$r = 1 \ cm   = 0.01 \ m$$
$$\therefore$$   $$F =\dfrac{4\pi\times 10^{-7}\times 20\times 75}{2\pi\times 0.01} =0.03 \ N/m$$

Let the length of wire Q be $$1 \ m$$  i.e.  $$l =1  \ m$$
So weight of wire Q   $$mg = 0.025\times 1 = 0.025 \ N$$
Since current in both the wire is flowing in opposite directions, so both wire will repel each other.
Force acting on the wire Q due to P    $$F = \dfrac{\mu_o I_1I_2 l }{2\pi r}$$
$$\therefore$$   $$F =\dfrac{4\pi\times 10^{-7}\times 50\times 25\times 1}{2\pi\times r} = \dfrac{25\times 10^{-5}}{r}$$
For the wire Q to be in equilibrium,    $$F = mg$$
$$\therefore$$   $$\dfrac{25\times 10^{-5}}{r} =0.025$$
$$\implies \ r = 0.01   \ m  =1 \ cm$$

Let, $$\overrightarrow {B} = B_1\hat{i}+B_2\hat{j}+B_3\hat{k}$$
Hence  $$q(v\hat{i}\times (B_1\hat{i}+B_2\hat{j}+B_3\hat{k})]= qvB (-\hat{j}+4\hat{k})$$
$$\Rightarrow B_2\hat{k} -B_3\hat{j} = -3B\hat{j}+4b\hat{k}$$
$$\Rightarrow B_2= 4B, B_3=3B$$
Also $$q[v\hat{j}\times (B_1\hat{i}+B_2\hat{j}+B_3\hat{k})]= qvB(3i)$$
$$\Rightarrow -B_2\hat{k} + B_3\hat{j} = 3B\hat{i}$$
$$\Rightarrow B_1=0 $$
Hence $$\overrightarrow {B} = 4B\hat{j}+3B\hat{k}$$
$$\hat{B} =\frac{\overrightarrow {B}}{|\overrightarrow {B}|}= \frac{4}{5}\hat{j}+\dfrac{3}{5}\hat{k}$$

dF (force of attraction) due to strip of thickness dx at a distance $$x dF =\dfrac{\mu_0iI_1}{2\pi x }\left( \dfrac{I_2dx}{b}\right) $$
$$F= \dfrac{\mu_0I_1I_2}{2\pi b} \int _a^{a+b}\dfrac{dx}{x}= \dfrac{\mu_0I_1I_2}{2\pi b} In \left(\dfrac{a+b}{b}\right)$$

Here, r, called the gyroradius or cyclotron radius, is the radius of curvature of the path of a charged particle with mass m and charge q, moving at a speed v perpendicular to a magnetic field of strength B. In other words, it is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field. If the velocity is not perpendicular to the magnetic field, then v is the component of the velocity perpendicular to the field. The component of the velocity parallel to the field is unaffected since the magnetic force is zero for motion parallel to the field. We’ll explore the consequences of this case in a later section on spiral motion

$$\dfrac{\mu_0 1 0 i}{2 \pi x} = \dfrac{\mu_0 i 40}{2 \pi (10 - x)}$$

The reason is magnetic flux is maximum through coil only if the face of coil is normal to the flux meaning the area vector of the face of coil must be parallel to the flux always , which you can understand from the following equation for torque i.e,

$$T=BINA\cos(a)$$ where a is the angle between area vector and magnetic flux , so you must have noticed that the T will be maximum only if $$\cos(a)$$ is $$1$$ i.e, parallel to flux 

The magnetic field present in the region affects the direction of motion of the particle but it doesn't affect the magnitude of the speed of the particles. So, it is fairly possible that the particle comes out with the same speed as the initial speed with which it enters the field.

$$v=3\times 10^6m/s$$  $$B=0.6T$$    $$m=1.67\times 10^{-27}kg$$

The deflection of the needle of galvanometer can be increased by:

We take a thick insulated copper wire and fix it in such a way that the portion $$AB$$ of the wire is in the north-south direction as shown in fig. A plotting compass $$M$$ is placed under the wire $$AB$$. The two ends of the wire are connected to a battery through a switch. When no current is flowing in the wire $$AB$$, the compass needle is parallel to the wire $$AB$$ and points in the usual north-south direction. When current is passed through wire $$AB$$ by closing the switch, we find that the compass needle is deflected from its north-south position. On opening the switch, the compass needle returns to its original position.
Thus, the deflection of compass needle by the current carrying wire shows that magnetic field is associated with an electric current.

When the south pole of the magnet is moved away, the galvanometer deflects to the left.

If the magnet is at rest in the coil, then their will be no deflection in the galvanometer. 

When the north pole is removed from the coil, the galvanometer deflects to the left.

The use of the stronger magnet increases the flow of the current.

In the presence of magnetic field; the force on a charged particle is given as:
$$F=q\times v \times B$$
Where $$q$$ is the charge of the particle
$$v$$ is the velocity of the particle
$$B$$ is the magnetic field.
According to question the direction of charged particle is clock wise and, the magnetic field is inwards ( perpendicular to the plane of paper in downwards direction).See the image attached.
Apply Fleming's left hand rule, according to the rule if you point your thumb, fore finger and central finger of left hand perpendicular to each other then
The thumb represents the force acting on direction of motion.
The fore finger represents the direction of magnetic field.
The central finger represents the direction of current i.e propagation of charges.

Strictly speaking electromagnetic induction is a process in which changing magnetic field produces electric current. Because electrons have their own magnetic field when an external changing magnetic field is applied to them they move. The movement of those electron produces current. Hence, we can say that the current will produce. But we won't be able to measure it or find out about it for instance as soon as we connect it to the galvanometer circuit becomes complete and that is not allowed according to the question.

Galvanometer pointer moves to other sides showing that the direction of momentarily indeed current has been reversed.
The phenomenon takin place here is electromagnetic induction. When the current passed through coil $$B$$ is stopped, the magnetic field linked with coil $$A$$ changes due to which an induced is produced in the coil.

Galvanometer pointer moves to one side for a moment showing that a momentary current is included in the coil $$A$$.

The galvanometer gives a reading to the left as the magnet moving away because the north pole goes away from the coil in this case.

The effect produced by the moving magnet is termed as Electromagnetic induction..

If there is no motion of the wire in the magnetic field, there will be no induced current in the wire.

The two changes that are observed ion the galvanometer reading are:

In a perpendicular magnetic field a charged particle traverses a circular path. 

Radius of circular path of a charged particle, $$r=\dfrac{mv}{qB}=\dfrac{P}{qB}$$
For same linear momentum P and magnetic field $$B$$,
$$r \alpha \dfrac{1}{q}$$
$$\therefore \dfrac{r_a}{r_p}=\dfrac{q_p}{q_a}=\dfrac{+e}{+2e}=\dfrac{1}{2}$$

(a) Deflection increases. 
(b) Direction of deflection is reversed. 
(c) Deflection decreases.

An iron strip can be magnetised with the help of a magnet by following method-

i. Take a wooden table and place the desired iron strip to be magnetised on it.
ii. Take a bar magnet in such a manner that one end of the magnet is held in the hands and the other end is on one edge of the strip.
iii. Move the magnet, without lifting, along the length of the strip till the other edge is reached.
iv. Repeat the above steps several times.
v. Take the iron fillings and extend it on the strip. If it gets attracted the strip gets magnetised and if it is not then repeat the steps few more times.

When we close the key, electric current start flowing in the circuit and due to this a magnetic force is created which deflects the compass’s needle.

$$\large\underline{\textbf{Concept applied:}}$$

(a) Figure shows that five conductors AA' , BB' ,CC' , DD' and EE' are along height of regular pentagonal prism ABCDE.

The magnitude of the magnetic field produced at the given point:
1. Magnetic field Increases as the current through wire increases.
2. Magnetic field Decreases as the distance from the wire increases.

Transmission of current on long distant can result in lot of loss of electrical energy as the energy get transmitted into heat energy. Whereas in case of AC current the loss of electrical energy is less as the wire gets less heated.

Magnetic lines of force are circular near the wire became parallel at the middle point of the coil.

Electric motor, electric generator, loud speaker and measuring instruments.

Given that

$$\textbf{magnetic filed}$$

When the proton moves parallel or anti parallel to the direction of magnetic field, proton does not get deflected.

When a current carrying conductor is placed in a magnetic field, it experiences a force.
When electric current flows through the conductor, a magnetic field is produced around it. During the interaction between the magnetic field of this conductor and the magnetic field of the magnet, the wire and magnet exert equal and opposite forces on each other. The magnet being heavy remains stationary and the wire moves due to this force.

AC generator: armature.

From the figure,
$$\sin \alpha = \dfrac {d}{R} = \dfrac {d\ q\ B}{mv}$$,
As radius of the arc $$R = \dfrac {mv}{qB}$$, where $$v$$ is the velocity of the particle, when it enteres into the field. From initial condition of the problem,
$$qV = \dfrac {1}{2} mv^{2}$$ or, $$v = \sqrt {\dfrac {2qV}{m}}$$
Hence, $$\sin \alpha = \dfrac {dqB}{m\sqrt {\dfrac {2qV}{m}}} = dB \sqrt {\dfrac {q}{2mV}}$$
and $$\alpha = \sin^{-1} \left (dB \sqrt {\dfrac {q}{2mV}}\right ) = 30^{\circ}$$, on putting the values.

(a) The total time of acceleration is,
$$t = \dfrac {1}{2v}\cdot n$$,
where $$n$$ is the number of passages of the Dees.
But, $$T = neV = \dfrac {B^{2} e^{2} r^{2}}{2m}$$
or, $$n = \dfrac {B^{2}er^{2}}{2mV}$$
So, $$t = \dfrac {\pi}{eB/m} \times \dfrac {B^{2}er^{2}}{2mV} = \dfrac {\pi B r^{2}}{2V} = \dfrac {\pi^{2} mv\ r^{2}}{eV} = 30\mu s$$
(b) The distance covered is, $$s = \displaystyle \sum v_{n}\cdot \dfrac {1}{2v}$$
But, $$v_{n} = \sqrt {\dfrac {2eV}{m}} \sqrt {n}$$,
So, $$s = \sqrt {\dfrac {eV}{2mv^{2}}} \displaystyle \sum \sqrt {2} = \sqrt {\dfrac {eV}{2mv^{2}}} \int \sqrt {n} dn = \sqrt {\dfrac {eV}{2mv^{2}}} \dfrac {2}{3} n^{3/2}$$
But, $$n = \dfrac {B^{2}e^{2}r^{2}}{2eVm} = \dfrac {2\pi^{2} mv^{2}r^{2}}{eV}$$
Thus, $$s = \dfrac {4\pi^{3}v^{2}mr^{2}}{3eV} = 1.24\ km$$.

When the metal ball moves in the magnetic field, the free electrons are distributed over the surface of the ball due to the action of the Lorentz force so that the resultant electric field in the bulk of the ball is uniform and compensates the action of the magnetic field. After the equilibrium has been attained, the motion of electrons in the bulk of the metal ceases. Therefore, the electric field strength is
$${ E }_{ res }q+q\left[ v\times B \right] =0\quad \quad $$
whence
$${ E }_{ res }=\left[ B\times v \right] $$
We arrive at the conclusion that the uniform electric field emerging in the bulk of the ball has the magnitude
$$\left| { E }_{ res } \right| =\left| B \right| \left| v \right| \sin { \alpha  } $$
The maximum potential difference $$\Delta { \phi  }_{ max }$$ emerging between the points on the ball diameter parallel to the vector $${E}_{res}$$ is
$$\Delta { \phi  }_{ max }=\left| { E }_{ res } \right| .2r=\left| B \right| \left| v \right| \sin { \alpha  } .2r$$

(a) For motion along a circle, the magnetic force acted on the particle, will provide the centripetal force, necessary for its circular motion.
i.e. $$\dfrac {mv^{2}}{R} = evB$$ or, $$v = \dfrac {eBR}{m}$$
and the period of revolution, $$T = \dfrac {2\pi}{\omega} = \dfrac {2\pi R}{v} = \dfrac {2\pi m}{eB}$$
(b) Generally, $$\dfrac {d\vec {p}}{dt} = \vec {F}$$
But, $$\dfrac {d\vec {p}}{dt} = \dfrac {d}{dt} \dfrac {m_{0}\vec {v}}{\sqrt {1 - (v^{2}/c^{2})}} = \dfrac {m_{0}\overset {\cdot}{\vec {v}}}{\sqrt {1 - (v^{2}/ c^{2})}} + \dfrac {m_{0}}{(1 - (v^{2}/ c^{2})^{3/2}} \dfrac {\vec {v}(\vec {v} \cdot \overset {\cdot}{\vec {v}})}{c^{2}}$$
For transverse motion, $$\vec {v} \cdot \overset {\cdot}{\vec {v}} = 0$$ so,
$$\dfrac {d\vec {p}}{dt} = \dfrac {m_{0}\overset {\cdot}{\vec {v}}}{\sqrt {1 - (v^{2}/ c^{2})}} = \dfrac {m_{0}}{\sqrt {1 - (v^{2}/ c^{2})}} \dfrac {v^{2}}{r}$$, here.
Thus, $$\dfrac {m_{0}v^{2}}{r\sqrt {1 - (v^{2}/c^{2})}} = Bev$$ or, $$\dfrac {v/c}{\sqrt{ 1 - (v^{2}/c^{2})}} = \dfrac {Ber}{m_{0}c}$$
or, $$\dfrac {v}{c} = \dfrac {Ber}{\sqrt {B^{2}e^{2}r^{2} + m_{0}^{2}c^{2}}}$$
Finally, $$T = \dfrac {2\pi r}{v} = \dfrac {2\pi m_{0}}{eB\sqrt {1 - v^{2}/c^{2}}} = \dfrac {2\pi}{cBe} \sqrt {B^{2} e^{2}r^{2} + m_{0}^{2}c^{2}}$$.

Using Fleming’s left hand rule we can easily find out that the nature of the charge on the particle is positive.

Given $$q={ 10 }^{ -3 }C;m={ 10 }^{ -3 }㎏$$
$$\vec { B } =-{ b }_{ 0 }x\hat { k } ;{ b }_{ 0 }=\cfrac { 1 }{ 10 } { T }/{ m }$$
$$v=20㎧$$
To find $${ x }_{ 0 }$$
Solution:-
Force acting on a charged particle moving in magnetic field$$=q\vec { v } \times \vec { B } $$
Magnetic force at orgin is in $$+y$$ direction. The particle moves in $$x-y$$ plane. The force is always perpendicular to the velocity so it will only change the direction of velocity vector but not hte magnitude. As magnetic field is non uniform its path is not a perfect circle. Let at point $$A(x,y,)$$ velocity makes angle $$\theta $$ with x-direction
$${ F }_{ m }=q{ v }_{ 0 }B\sin { 90 } =q{ v }_{ 0 }{ b }_{ 0 }x\rightarrow 1$$
Velocity along x direction $${ v }_{ x }=\cfrac { dx }{ dt } ={ v }_{ 0 }\cos { \theta  } \rightarrow 2$$
Velocity along y direction$${ v }_{ y }=\cfrac { dy }{ dt } =v\sin { \theta  } $$
Now,
$${ F }_{ y }={ F }_{ m }\cos { \theta  } $$
$$\Rightarrow { a }_{ y }=\cfrac { { F }_{ m } }{ m } \cos { \theta  } $$
$$\Rightarrow \cfrac { d{ v }_{ y } }{ dt } =\cfrac { { b }_{ 0 }x }{ m } \times q{ v }_{ 0 }\times \cos { \theta  } $$
$$\Rightarrow \cfrac { d{ v }_{ y } }{ dx } \cfrac { dx }{ dt } =\cfrac { { b }_{ 0 }x }{ m } q\left( { v }_{ 0 }\cos { \theta  }  \right) $$
$$\Rightarrow \cfrac { d{ v }_{ y } }{ dx } \times { v }_{ 0 }\cos { \theta  } =\cfrac { { b }_{ 0 }x }{ m } q\left( { v }_{ 0 }\cos { \theta  }  \right) $$(from equation $$2$$)
$$\Rightarrow d{ v }_{ y }=\left( \cfrac { { b }_{ 0 }x }{ m }  \right) xdx$$
At maximum displacement along x direction velocity of particle is along y direction i.e., $${ v }_{ y }={ v }_{ 0 }$$
$$\Rightarrow \int _{ 0 }^{ { v }_{ 0 } }{ d{ v }_{ y } } =\cfrac { { b }_{ 0 }q }{ m } \int _{ 0 }^{ { x }_{ max } }{ xdx } $$
$$\Rightarrow { v }_{ 0 }=\cfrac { { b }_{ 0 }q }{ m } \cfrac { { x }_{ max }^{ 2 } }{ 2 } $$
$$\Rightarrow { x }_{ max }=\sqrt { \cfrac { 2{ v }_{ 0 }m }{ q{ b }_{ 0 } }  } $$
$$\Rightarrow { x }_{ max }=\sqrt { \cfrac { 2\times 20\times { 10 }^{ -3 } }{ { 10 }^{ -3 }\times { 1 }/{ 10 } }  } =\sqrt { { \left( 2 \right)  }^{ 2 }\times { \left( 10 \right)  }^{ 2 } } $$
$$\Rightarrow { x }_{ max }=20m=20\times 100㎝$$
$$\Rightarrow { x }_{ max }=4{ x }_{ 0 }420\times 100㎝$$
$$\Rightarrow { x }_{ 0 }=5\times 100㎝$$
$$\Rightarrow { x }_{ 0 }=5m$$

In three dimensional co-ordinates system, axes are assumed according to right hand screw rule. Consider such a system shown in fig. In this system positive z direction is normal to plane of paper and is directed toward the reader.
The particle is positively charged, centre of its circular path is at origin and when the particle is on positive y-axis, its velocity is directed along positive x-direction, it means that the particle is moving clockwise in fig.
Since the particle is positively charged, therefore, current associated with its motion is also clockwise or along positive x direction at (0, 0.4, 0).
Lorentz's force is toward origin. Therefore, according to Fleming's left hand rule, magnetic induction B must be along positive z-direction.
Radius of circular path followed by this particle is $$r_1=0.4 m$$
But $$r_1=\dfrac {m_1v_1}{q_1B}\Rightarrow B=\dfrac {m_1v_1}{q_1r_1}=0.5 T$$
Now the second particle collides and gets struck with it. Velocity of combined particle can be found out by applying law of conservation of momentum.
$$(m_1+m_2)\vec v=m_1\vec v_1+m_2\vec v_2$$
$$(40\times 10^{-3}+10\times 10^{-3})\vec v$$$$=(40\times 10^{-3)}\times 5\hat i+(10\times 10^{-3})\dfrac {40}{\pi}\hat k$$
$$\therefore$$ x-direction of velocity of combined particle is $$v_x=4 ms^{-1}$$ and z-direction, $$v_z=\frac {8}{\pi}ms^{-1}$$
Due to $$v_x$$, combined particle tries to move clockwise along a circular path and due to $$v_z$$ it ties to move uniformly along z-axis.
Therefore, its ultimate path becomes helix. 

Let at any instant velocity of charge is $$v$$, then force on charge due to magnetic field caused by wire is given by
$$m{ a }_{ x }\hat { i } +m{ a }_{ y }\hat { j } =q({ v }_{ x }\hat { i } +{ v }_{ z }\hat { k } )\times \cfrac { { \mu  }_{ 0 }I }{ 2\pi x } \hat { j } $$
$$\Rightarrow m{ a }_{ x }\hat { i } +m{ a }_{ y }\hat { j } =q{ v }_{ x }\cfrac { { \mu  }_{ 0 }I }{ 2\pi x } \hat { k } -q{ v }_{ z }\cfrac { { \mu  }_{ 0 }I }{ 2\pi x } \hat { i } $$
$$\Rightarrow m{ a }_{ x }=-\cfrac { { q\mu  }_{ 0 }I }{ 2\pi  } \cfrac { { v }_{ z } }{ x } $$
$$\Rightarrow { a }_{ x }=-\cfrac { { q\mu  }_{ 0 }I }{ 2\pi m } \cfrac { { v }_{ z } }{ x } $$

Time taken by particle to reach the screen: $$t=\dfrac {l}{vcos \alpha}$$
Radius of helical path: $$R=\dfrac {mv sin\alpha}{qB}$$
Angle rotated in time: $$\theta=\omega t=\dfrac {qB}{m}t$$
Required distance:
$$r=2R sin\left (\dfrac {\theta}{2}\right )=\dfrac {2 mv sin\alpha}{qB}sin \left (\dfrac {qB}{2m}\dfrac {1}{v cos\alpha}\right )$$

Magnetic field  $$B =1\times 10^{-4} \ Wb/m^2$$
Mass of electron  $$m = 9.1\times 10^{-31} \ kg$$
Magnitude of charge of electron  $$e = 1.6\times 10^{-19} \ C$$
Kinetic energy of electron  $$K =10eV$$
(i) : Speed of electron   $$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2\times  10\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}} =5.93\times 10^6 \ m/s$$
(ii) :  Radius of circular path    $$R = \dfrac{mv}{eB}$$
$$\implies \ R = \dfrac{9.1\times 10^{-31}\times 5.93\times 10^6 }{1.6\times 10^{-19}\times 10^{-4}} = 33.73\times 10^{-2} \ m = 33.73 \ cm$$
(iii) :  Time period   $$T = \dfrac{2\pi m}{eB}$$
$$\implies \ T = \dfrac{2\pi\times 9.1\times 10^{-31}}{1.6\times 10^{-19}\times 10^{-4}} = 35.72\times 10^{-8} \ s$$

Let length of solenoid $$=l$$

(a)

Force on one wire due to other is $$F=BI_2l$$ where magnetic field $$B=\dfrac{\mu_0I_1}{2\pi r}$$  due to first wire at distance $$r$$.

(a)Force on proton$$=q\overrightarrow { V } \times \overrightarrow { B } $$ ($$v$$=velocity of proton)

Motion of a charged particle in a uniform magnetic field : 

Radius of circular path  $$R = \dfrac{mv}{qB}$$
Given:  $$v = 10^4 \ m/s$$,    $$m = 3.32\times 10^{27} \ kg$$,    $$q = 1.6\times 10^{-19}\ C$$  and  $$B = 10^{-3} \ T$$
$$\therefore \ R  = \dfrac{(3.32\times 10^{-27})\times (10^4)}{(1.6\times 10^{-19})\times 10^{-3}}$$
$$\implies \ R = 0.2 \ m$$

The force acting on the charged particle placed in a magnetic field is given as

$$F = q\left( {v \times B} \right)$$

If the velocity is perpendicular to that of the magnetic field then the force acting on the charged particle is perpendicular to that of velocity and magnetic field. There will not be any components of the force. This is possible only when the trajectory of the field is circular.

The particles of mass is m and charge q is moving with a uniform velocity v. This particle is entering in a magnetic field. Then the radius of circular arc described by particle in magnetic field is

(a)

$$ qvB=\dfrac{m{{v}^{2}}}{r} $$

$$ r=\dfrac{mv}{qB} $$

(b) From the figure,

$$ \angle MAO=90 $$

$$ \angle NAC=90 $$

$$ \angle AOC=180-\left( \dfrac{\pi }{4}+\dfrac{\pi }{4} \right) $$

$$ =90 $$

(c) Distance covered by particle inside magnetic field

$$l=r\theta$$

$$ l=\dfrac{mv}{qB}\times \dfrac{\pi }{4} $$

$$ =\dfrac{\pi mv}{4qB} $$

$$ t=\dfrac{l}{v}=\dfrac{\pi m}{4qB} $$

(d) if charge is negative then

$$r=\dfrac{mv}{qB}$$

Angle subtended will be

$$ 180+\left( \dfrac{\pi }{4}+\dfrac{\pi }{4} \right) $$

$$ =270 $$

$$t=\dfrac{3\pi m}{2qB}$$

 When a current carrying coil is suspended in a uniform magnetic field it is acted upon by a torque.

We take the current $$(i = 50 A)$$ to flow in the $$+x$$ direction, and the electron to be at a point $$P$$, which is $$r = 0.050m$$ above the wire (where "up" is the $$+y$$ direction). thus, the field produced by the current points in the $$+z$$ direction at $$P$$. we obtain

The period of revolution for the iodine ion is $$T=\dfrac{2 \pi r}{v} = \dfrac{2 \pi m}{Bq}$$, which gives

$$m = \dfrac{BqT}{2\pi}$$

$$=\dfrac{(45.0 \times 10^{-3}T)(1.60 \times 10^{-19}C)(1.29 \times 10^{-3}s)}{(7)(2\pi)(1.66 \times 10^{-27}kg/u)}$$

$$ =127u$$.

Magnetic forces per unit length on wire $$(2)$$
$$F=\dfrac {\mu_0}{2\pi r}. 3I^2 -\dfrac {\mu_0 12I^2}{2\pi (2r)}$$
$$=\dfrac {3}{2}\dfrac {\mu_0 I^2}{\pi r}-3\dfrac {\mu_0 I^2}{\pi r}=-\dfrac 32 \dfrac {\mu_0 I^2}{\pi r}$$
Hence, $$F=\dfrac 32 \dfrac {\mu_0 I^2}{\pi r}$$ in the direction of wire $$1$$.

Ans: Positive Charge

Choose the wire along the z-axis, and the initial direction of the electron, along the x-axis.
Then the magnetic field in the $$x-z$$ plane is along the y-axis and outside the wire it is,
$$B = B_{y} = \dfrac {\mu_{0}I}{2\pi x}, (B_{x} = B_{z} = 0$$, if $$y = 0)$$
The motion must be confined to the $$x - z$$ plane. Then the equations of motion are,
$$\dfrac {d}{dt}mv_{x} = -ev_{z} B_{y}$$
$$\dfrac {d(mv_{z})}{dt} = +ev_{x} B_{y}$$
Multiplying the first equation by $$v_{x}$$ and the second by $$v_{z}$$ and then adding,
$$v_{x} \dfrac {dv_{x}}{dt} + v_{z} \dfrac {dv_{z}}{dt} = 0$$
or, $$v_{x}{2} + v_{z}^{2} = v_{0}^{2}$$, say, $$v_{z} = \sqrt {v_{0}^{2} - v_{x}^{2}}$$
Then, $$v_{x} \dfrac {dv_{x}}{dx} = -\dfrac {e}{m} \sqrt {v_{0}^{2} - v_{x}^{2}} \dfrac {\mu_{0}I}{2\pi x}$$
or, $$-\dfrac {v_{x}dv_{x}}{\sqrt {v_{0}^{2} - v_{x}^{2}}} = \dfrac {\mu_{0} Ie}{2\pi m} \dfrac {dx}{x}$$
Integrating, $$\sqrt {v_{0}^{2} - v_{x}^{2}} = \dfrac {\mu_{0} Ie}{2\pi m} ln \dfrac {x}{a}$$
on using, $$v_{x} = v_{0}$$, if $$x = a$$ (i.e. initially).
Now, $$v_{x} = 0$$, when $$x = x_{m}$$,
so, $$x_{m} = ae^{v_{0}/b}$$, where $$b = \dfrac {\mu_{0} Ie}{2\pi m}$$.

Let us take the point $$A$$ as the origin $$O$$ and the axis of the solenoid as z-axis. At an arbitrary moment of time let us resolve the velocity of electron into its two rectangular components, $$\vec {v}_{\parallel}$$ along the axis and $$\vec {v}_{\perp}$$ to the axis of solenoid. We know the magnetic remain constant in the $$x-y$$ plane. Thus $$\vec {v}_{\perp}$$ can change only its direction as shown in the Fig.. $$\vec {v}_{\parallel}$$ will remain constant as it is parallel to $$\vec {B}$$.
Thus at $$t = t$$
$$v_{x} = v_{\perp} \cos \omega t = v\sin \alpha \cos \omega t$$,
$$v_{y} = v_{\perp} \sin \omega t = v\sin \alpha \sin \omega t$$
and $$v_{z} = v\cos \alpha$$, where $$\omega = \dfrac {eB}{m}$$
As at $$t = 0$$, we have $$x = y = z = 0$$, so the motion law of the electron is.
$$\left.\begin{matrix}z = v\cos \alpha t\\ x = \dfrac {v\sin \alpha}{\omega} \sin \omega t\\ y = \dfrac {v\sin \alpha}{\omega} (\cos \omega t - 1)
\end{matrix}\right|$$
(The equation of the helix)
On the screen, $$z = l$$, so $$t = \dfrac {l}{v\cos \alpha}$$.
Then, $$r^{2} = x^{2} + y^{2} = \dfrac {2v^{2} \sin^{2}\alpha}{\omega^{2}} \left (1 - \cos \dfrac {\omega l}{v\cos \alpha}\right )$$
$$r = \dfrac {2v\sin \alpha}{\omega} \left |\sin \dfrac {\omega l}{2v\cos \alpha}\right | = 2\dfrac {mv}{eB}\sin \alpha \left |\sin \dfrac {leB}{2mv\cos \alpha}\right |$$.

On the one hand,
$$\dfrac {dp}{dt} = eE = \dfrac {e}{2\pi r} \dfrac {d\Phi}{dt} = \dfrac {e}{2\pi r} \dfrac {d}{dt}\int_{0}^{r} 2\pi r' B(r') dr'$$
On the other,
$$p = B(r) er, r = constant$$.
so, $$\dfrac {dp}{dt} = er \dfrac {d}{dt} B(r) \equiv er\overset {\cdot}{B}(r)$$
Hence, $$er\overset {\cdot}{B}(r) = \dfrac {e}{2\pi r} \pi r^{2} \dfrac {d}{dt} < B >$$
So, $$\overset {\cdot}{B}(r) = \dfrac {1}{2} \dfrac {d}{dt} < B >$$
This equation is most easily satisfied by taking $$B(r_{0}) = \dfrac {1}{2} < B >$$.

$$\displaystyle We\ know,\ \frac{mv^{2}}{r} =qvB$$
$$\displaystyle or,\ r=\frac{\sqrt{2mE}}{qB} \ \ \ \ \because E=\frac{1}{2} mv^{2}$$
$$\displaystyle or,\ r_{p} =\frac{\sqrt{2m_{p} E}}{q_{p} B} \ \ \ \ and\ r_{\alpha } =\frac{\sqrt{2m_{\alpha } E}}{q_{\alpha } B} \ $$
$$\displaystyle \ \frac{r_{p}}{r_{\alpha }} =\frac{\sqrt{2m_{p} E}}{q_{p} B} \ \ \times \ \ \frac{q_{\alpha } B}{\sqrt{2m_{\alpha } E}} \ $$
$$\displaystyle \ \frac{r_{p}}{r_{\alpha }} =\sqrt{\frac{m_{p}}{4m_{p}}} \times \frac{2q_{p}}{q_{p}} \ $$
$$\displaystyle \ \frac{r_{p}}{r_{\alpha }} =1\ $$

Hint - Two parallel current-carrying wires will exert an attractive force on each other, if their currents are in the same direction. If the parallel wires currents in opposite directions, the wires repel each other.
Step-1 Given data
Step-2 Calculation of forces on wires AB and CD.