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A Monoenergetic Electron Beam with Electron Speed of 5.20 × 106 M S−1 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 × 1011 - Physics

A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the a radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1.

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Solution

Speed of an electron, v = 5.20 × 106 m/s

Magnetic field experienced by the electron, B = 1.30 × 10−4 T

Specific charge of an electron, e/m = 1.76 × 1011 C kg−1

Where,

e = Charge on the electron = 1.6 × 10−19 C

m = Mass of the electron = 9.1 × 10−31 kg−1

The force exerted on the electron is given as:

`F = e|vecv xx vecB|`

`= evBsin theta`

θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

`:. theta  = 90^@`

F = eVB   ...(1)

The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force `(F = (mv^2)/r)` for the beam.

Hence, equation (1) reduces to:

`evB = (mv^2)/r`

`:. r = mv/(eB) = v/((e/m)B)`

`= (5.20 xx 10^6) /(1.76 xx 10^11) xx 1.30 xx    10^(-4) = 0.227 m = 22.7 cm`

Therefore, the radius of the circular path is 22.7 cm.

Concept: Photoelectric Effect and Wave Theory of Light
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 21.1 | Page 409
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