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 $4.8 \times 10^{6} m s^{- 1}$ normal to the field. Explain why the path of the electron is a circle?
Determine the radius of the circular orbit. $(e = 1.5 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g)$ .

Also obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

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Solution

At every instant, the direction of motion of the particle is perpendicular to the direction of force. Hence the path of the electron is circular.
We know that, at equilibrium,
$e v B = m v^{2} / r$
also, $v = r ω$
$\Rightarrow f = B e / 2 π m$
$f = 18 M H z$
Frequency is independent of velocity.

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(1 G = 10^{- 4} T)

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4.8 \times 10^{6} m s^{- 1}

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(e = 1.5 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g)

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At every instant, the direction of motion of the particle is perpendicular to the direction of force. Hence the path of the electron is circular.We know that, at equilibrium,evB=mv2/ralso, v=rω⇒f=Be/2πmf=18 MHzFrequency is independent of velocity.

At every instant, the direction of motion of the particle is perpendicular to the direction of force. Hence the path of the electron is circular.
We know that, at equilibrium,
$e v B = m v^{2} / r$
also, $v = r ω$
$\Rightarrow f = B e / 2 π m$
$f = 18 M H z$
Frequency is independent of velocity.