Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

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Q. 389. A non-relativistic proton beam passes without deviation through the region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with E = 120 kV/m and B = 50 mT. Then the beam strikes a grounded target. Find the force with which the beam acts on the target if the beam current is equal to I = 0.80 mA.

Solution. 389. In crossed field,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Then,F = force exerted on the plate  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 390. Non-relativistic protons move rectilinearly in the region of space where there are uniform mutually perpendicular electric and magnetic fields with E = 4.0 kV/m and B = 50 mT. The trajectory of the protons lies in the plane xz (Fig. 3.102) and forms an angle φ = 30° with the x axis. Find the pitch of the helical trajectory along which the protons will move after the electric field is switched off. 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Solution. 390. When the electric field is switched off, the path followed by the particle will be helical, and pitch, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev is the velocity of the particle, parallel to  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev the  time period of revolution.)

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev     (1)

Now, when both the fields were present,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev as no net force was effective

or,    Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev    (2)

From (1) and (2), Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 391. A beam of non-relativistic charged particles moves without deviation through the region of space A (Fig. 3.103) where there are transverse mutually perpendicular electric and magnetic fields with strength E and induction B. When the magnetic field is switched off, the trace of the beam on the screen S shifts by Δx. Knowing the distances a and b, find the specific charge q/m of the particles. 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Solution. 391. When there is no deviation,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or, in scalar from,Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev   (1)

Now, when the magnetic field is switched on, let the deviation in the field be x. Then, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

where t is the time required to pass through this region, 

also,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev    (2)

For the region where the field is absent, velocity in upward direction

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev   (3)

Now,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev  (4)

From (2) and (4),

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 392. A particle with specific charge qim moves in the region of space where there are uniform mutually perpendicular electric and magnetic fields with strength E and induc- tion B (Fig. 3.104). At the moment t = 0 the particle was located at the point O and had zero velocity. For the non-relativistic case find:
 (a) the law of motion x (t) and y (t) of the particle; the shape of the trajectory;
 (b) the length of the segment of the trajectory between two nearest points at which the velocity of the particle turns into zero;
 (c) the mean value of the particle's velocity vector projection on the x axis (the drift velocity). 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Solution. 392. (a) The equation of motion is, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Now,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

So, the equation becomes, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Here,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev The last equation is easy to integrate; 

vz = constant = 0, 

since vz is zero initially. Thus integrating again,

z = constant = 0,

and motion is confined to the x - y plane. We now multiply the second equation by i and add to the first equation.

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

we get the equation, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

This equation after being multiplied by Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev be rewritten as,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

and integrated at once to give, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

where C and α are two real constants. Taking real and imaginary parts.

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Since vy = 0, when t = 0, we can take α = 0, then vx = 0 at t = 0 gives, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev and we get,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Integrating again and using x = y = 0, at t = 0, we get

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

This is the equation of a cycloid.

(b) The velocity is zero, when ωt - 2nπ. We see that

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

The quantity inside the modulus is positive for 0 < ωt < 2 π. Thus we can drop the modulus and write for the distance traversed between two successive zeroes of velocity.

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Putting  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

(c) The drift velocity is in the x-direction and has the magnitude,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 393. A system consists of a long cylindrical anode of radius a and a coaxial cylindrical cathode of radius b (b < a). A filament located along the axis of the system carries a heating current I producing a magnetic field in the surrounding space. Find the least potential difference between the cathode and anode at which the thermal electrons leaving the cathode without initial velocity start reaching the anode.

Solution. 393. When a current I flows along the a x is, a magnetic field Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev is set up where  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev In terms of components,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Suppose a p.d. V is set up between the inner cathode and the outer anode. This means a potential function of the form 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

as one can check by solving Laplace equation. The electric field corresponding to this is,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

The equations of motion are,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

and Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev is the charge on the electron, 

Integrating the last equation, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

since vz = 0 where p = a. We now substitute this  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev in the other two equations to get

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Integrating and using v2 = 0, at p = b, we get,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

The RHS must be positive, for all a > p > b. The condition for this is,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 394. Magnetron is a device consisting of a filament of radius a and a coaxial cylindrical anode of radius b which are located in a uniform magnetic field parallel to the filament. An accelerating potential difference V is applied between the filament and the anode. Find the value of magnetic induction at which the electrons leaving the filament with zero velocity reach the anode.

Solution. 394. This differs from the previous problem in Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev and the magnetic field is along the z-direction. Thus Bx = By = 0, Bz = B

Assuming as usual the charge of the electron to be - | e |, w e write the equation of motion

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

and  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

The motion is confined to the plane z = 0. Eliminating B from the first two equations,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

so, as expected, since magnetic forces do not work,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

On the other hand, eliminating V, we also get,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

i.e.  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

The constant is easily evaluated, since v is zero at p = a. Thus,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

At  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

 or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 395. A charged particle with specific charge qim starts moving in the region of space where there are uniform mutually perpendicular electric and magnetic fields. The magnetic field is constant and has an induction B while the strength of the electric field varies with time as E = Em cos ωt, where ω = qB/m. For the non-relativistic case find the law of motion x (t) and y (t) of the particle if at the moment t = 0 it was located at the point O (see Fig. 3.104). What is the approximate shape of the trajectory of the particle? 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Solution. 395. The equations are as in Q.392.

with  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or multiplying by Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or integrating,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

since  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Integrating again, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

where  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev and we have used x = y = 0, at t = 0.


The trajectory is an unwinding spiral.


Q. 396. The cyclotron's oscillator frequency is equal to v = 10 MHz. Find the effective accelerating voltage applied across the dees of that cyclotron if the distance between the neighbouring trajectories of protons is not less than Δr = 1.0 cm, with the trajectory radius being equal to r = 0.5 m. 

Solution. 396. We know that for a charged particle (proton) in a magnetic field,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

But, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

So,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

On the other hand ΔE = 2 eV, where V is the effective acceleration voltage, across the Dees, there being two crossings per revolution. So,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 397. Protons are accelerated in a cyclotron so that the maximum curvature radius of their trajectory is equal to r = 50 cm. Find:
 (a) the kinetic energy of the protons when the acceleration is completed if the magnetic induction in the cyclotron is B = 1.0 T;
 (b) the minimum frequency of the cyclotron's oscillator at which the kinetic energy of the protons amounts to T = 20 MeV by the end of acceleration. 

Solution. 397. Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev Bev, or, mv = Ber

and Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

we get,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 398. Singly charged ions He+ are accelerated in a cyclotron so that their maximum orbital radius is r = 60 cm. The frequency of a cyclotron's oscillator is equal to v = 10.0 MHz, the effective accelerating voltage across the dees is V = 50 kV. Neglecting the gap between the dees, find:
 (a) the total time of acceleration of the ion;
 (b) the approximate distance covered by the ion in the process of its acceleration.

Solution. 398. (a) The total time of acceleration is,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

where n is the number of passages of the Dees.

But,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

So,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

(b) The distance covered is,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

But,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

So,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

But,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 399. Since the period of revolution of electrons in a uniform magnetic field rapidly increases with the growth of energy, a cyclotron is unsuitable for their acceleration. This drawback is rectified in a microtron (Fig. 3.105) in which a change ΔT in the period of revolution of an electron is made multiple with the period of accelerating field T0. How many times has an electron to cross the accelerating gap of a microtron to acquire an energy W = 4.6 MeV if ΔT = T0, the magnetic induction is equal to B = 107 mT, and the frequency of accelerating field to v = 3000 MHz? 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Solution. 399. In the nth orbit,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev We ignore the rest mass of the electron and write

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 400. The ill effects associated with the variation of the period of revolution of the particle in a cyclotron due to the increase of its energy are eliminated by slow monitoring (modulating) the frequency of accelerating field. According to what law ω (t) should this frequency be monitored if the magnetic induction is equal to B and the particle acquires an energy ΔW per revolution? The charge of the particle is q and its mass is m.

Solution. 400. The basic condition is the relativistic equation,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Or calling,   Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

we get,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

is the radius of the instantaneous orbit.  

The time of acceleration is,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

N is the number of crossing of either Dee.

But, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev there being two crossings of the Dees per revolution.

So, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Also,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Hence finally, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 401. A particle with specific charge q/m is located inside a round solenoid at a distance r from its axis. With the current switched into the winding, the magnetic induction of the field generated by the solenoid amounts to B. Find the velocity of the particle and the curvature radius of its trajectory, assuming that during the increase of current flowing in the solenoid the particle shifts by a negligible distance.

Solution. 401. When the magnetic field is being set up in the solenoid, and electric field will be induced in it, this will accelerate the charged particle. If B is the rate, at which the magnetic field is increasing, then.

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

After the field is set up, the particle will execute a circular motion of radius p, where 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 402. In a betatron the magnetic flux across an equilibrium orbit of radius r = 25 cm grows during the acceleration time at practically constant rate Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev Wb/s. In the process, the electrons acquire an energy W = 25 MeV. Find the number of revolutions made by the electron during the acceleration time and the corresponding distance covered by it.

Solution. 402. The increment in energy per revolution is Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev so the number of revolutions is,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

The distance traversed is, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 403. Demonstrate that electrons move in a betatron along a round orbit of constant radius provided the magnetic induction on the orbit is equal to half the mean value of that inside the orbit (the betatron condition).

Solution. 403. On the one hand,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

On the other

p = B (r) er, r = constant.

so, Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Hence,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

So,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

This equations is most easily satisfied by taking  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 404. Using the betatron condition, find the radius of a round orbit of an electron if the magnetic induction is known as a function of distance r from the axis of the field. Examine this problem for the specific case B = B0  — ar2, where B0 and a are positive constants. 

Solution. 404. The condition  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

This gives r0

In the present case,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

or,   Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 405. Using the betatron condition, demonstrate that the strength of the eddy-current field has the extremum magnitude on an equilibrium orbit. 

Solution. 405. The induced electric field (or eddy current field) is given by,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Hence,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

This vanishes for r = r0 by the betatron condition, where r0 is the radius of the equilibrium orbit


Q. 406. In a betatron the magnetic induction on an equilibrium orbit with radius r = 20 cm varies during a time interval Δt = 1.0 ms at practically constant rate from zero to B = 0.40 T. Find the energy acquired by the electron per revolution. 

Solution. 406. From the betatron condition,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

Thus,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

and  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

So, energy increment per revolution is, 

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev


Q. 407. The magnetic induction in a betatron on an equilibrium orbit of radius r varies during the acceleration time at practically constant rate from zero to B. Assuming the initial velocity of the electron to be equal to zero, find:
 (a) the energy acquired by the electron during the acceleration time;
 (b) the corresponding distance covered by the electron if the acceleration time is equal to Δt. 

Solution. 407. (a) Even in the relativistic case, we know that : p = Her

Thus,  Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

(b) The distance traversed is,

Irodov Solutions: Motion of Charged Particles In Electric And Magnetic Fields- 2 Notes | EduRev

on using the result of the previous problem.

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