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In a Chamber, a Uniform Magnetic Field of 6.5 G (1 G = 10^–4 T) is Maintained. an Electron is Shot into the Field with a Speed of 4.8 × 106 M S–1 Normal to the Field. Explain Why the Path of the Electron is a Circle. Determine the Radius of the Circular Orbit. - Physics

In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (= 1.6 × 10–19 C, me= 9.1×10–31 kg)

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Solution

Magnetic field strength, B = 6.5 G = 6.5 × 10–4 T

Speed of the electron, v = 4.8 × 106 m/s

Charge on the electron, e = 1.6 × 10–19 C

Mass of the electron, me = 9.1 × 10–31 kg

Angle between the shot electron and magnetic field, θ = 90°

Magnetic force exerted on the electron in the magnetic field is given as:

F = evB sinθ

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron,

`F_c=(mv^2)/r`

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

`F_c=F`

`(mv^2)/r=evBsintheta`

`r=(mv)/(Besin theta)`

`=(9.1xx10^-31xx4.8xx10^6)/(6.5xx10^-4xx1.6xx10^-19xxsin90^@)`

`=4.2xx10^-2m=4.2 cm`

Hence, the radius of the circular orbit of the electron is 4.2 cm.

 

Concept: Torque on Current Loop, Magnetic Dipole - Torque on a Rectangular Current Loop in a Uniform Magnetic Field
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 4 Moving Charges and Magnetism
Q 11 | Page 169
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