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#### (a) A monoenergetic electron beam with electron speed of 5.20*10^6 m/s is subject to a magnetic field of 1.30 *10^{-4} T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 *10^{11} C/kg. (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

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[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.] Qalaxia Knowlege Bot
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TOWARDS MULTI-WAVELENGTH OBSERVATIONS OF ...

4.3 Poynting Flux, GEM and LEM Variation with s in Self-Similar Model . . . . . . . . 47. 4.4 Streamplot of Magnetic Field between 10M and 100M in Self-Similar ... Qalaxia Master Bot
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I found an answer from www.toppr.com

Motion in Combined Electric and Magnetic Field: What is Magnetic ...

Magnetic fields are used to guide the motion of charged particles in ... The Van Allen radiation belts in space around the earth consist of these energetic charges trapped in ... to a uniform magnetic field B (no E) moves in a circular path with the radius ... force, q v × B, acts as a centripetal force and produces a circular motion  ...

For more information, see Motion in Combined Electric and Magnetic Field: What is Magnetic ... Veda
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a)

Given that

A monoenergetic electron beam speed v = 5.20 * 10^{6} m/s

Magnetic field B = 1.30 * 10^{-4} T

Specific charge of electron \frac{e}{m} = 1.76*10^{11} C/kg

Magnetic field is perpendicular to the beam velocity

So \theta = 90 \degree

Step 1: Set up an equation for the radius of the circular path of charged particle

Formulas required

Force acting on electron in magnetic field F = evB \sin \theta

Where, e - charge, B - magnetic field, v - velocity, \theta - angle between the velocity and the magnetic field.

F = evB   .........................(1)                        \because    \theta = 90 \degree

A charged particle moving in a circular path, so it acquires centripetal force

Centripetal force   F = \frac{mv^2}{r} ...............(2)

Where,  r - radius of the circular path

From equation(1) and (2) we can write as follows

evB = \frac{mv^2}{r}

r = \frac{mv^2}{evB}

r = \frac{mv}{eB}

Hence,  radius of the circular path of charged particle r = \frac{mv}{qB} ................(3)

Step 2: Plug in the given values in the equation (3) for radius of circular path

r = \frac{v}{\frac{e}{m} B}

r = \frac{5.20 * 10^{6}}{1.76*10^{11} * 1.30 * 10^{-4}}

r = 0.227 m

r = 22.7 cm

Therefore, radius of the circular path of charged particle   r = 22.7 cm

b)

Given that

Energy of the electron beam E = 20 MeV = 20* 10^{6} eV

E = 20 * 10^{6} * 1.6 * 10^{-19} J

Step 1: Set up an equation for the velocity of the electron

Kinematic energy K.E = \frac{1}{2} mv^2.

v^2 = \frac{2 K.E}{m}

v = \sqrt{\frac{2 K.E}{m}}

Step 2: Substitute the given values in the velocity equation

Mass of an electron m = 9.1 * 10^{-31} kg

v = \frac{2 * 20 * 10^{6} * 1.6 * 10^{-19}}{9.1 * 10^{-31}}

v = 2.65 * 10^{9} m/s

Step 3: Observation

Speed of light c = 38 10^{8} m/s

Speed of electron is greater than the speed of the light which is practically not possible.

K.E = \frac{1}{2} mv^2.   is valid only for v << c

So modify the formula of radius by replacing the relativistic mass

Modified mass m=m_0*(1-\frac{v^2}{c^2})^{-\frac{1}{2}}

m_o - mass of particle at rest position

Thus, valid formula for calculating the radius of circular path is as follows