In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^–4T) is maintained. An electron is shot into the field with a speed of 4.8 x 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. (e = 1.6 x 10^–31 C, m_e = 9.1 x 10^–31 kg)

Solution

Here given,
Uniform magnetic field, B = 6.5 x 10^–4 T
Speed with which the electron is shot, v = 4.8 x 10⁶ m/s
Charge on electron, e = 1.6 x 10^–19 C
Angle made by the electron with the field, θ = 90° Mass of electron, m = 9.1 x 10^–31 kg
Radius of the circular orbit, r =?

(i) Force on the moving electron due to magnetic field will be,

F = evB \sin θ .

The direction of this force is perpendicular to

\vec{v}

and

\vec{B}

therefore, this force will only change the direction of motion of the electron without affecting its velocity i.e., this force will provide the centripetal force to the moving electron and hence, the electron will move on the circular path.

If r is the radius of circular path traced by electron, then

e v B s i n 90 ° = m v^{2} / r \Rightarrow r = \frac{m v}{B e} \Rightarrow r = \frac{(9.1 \times 10^{- 31}) \times (4.8 \times 10^{6})}{(6.5 \times 10^{- 4}) \times (1.6 \times 10^{- 19})} = 4.2 \times 10^{- 2} m = 4.2 cm

Some More Questions From Moving Charges And Magnetism Chapter

Answer the following questions:
A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

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