4.11) In a chamber, a uniform magnetic field of  6.5 G ( 1G = 10 ^{-4}T ) is maintained. An electron is shot into the field with a speed of 4.8 \times 10 ^ 6 ms ^{-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} kg)

Answers (1)
S Sayak

The magnetic force on a moving charged particle in a magnetic field is given by \vec{F_{B}}=q\vec{V}\times \vec{B}

Since the velocity of the shot electron is perpendicular to the magnetic field, there is no component of velocity along the magnetic field and therefore the only force on the electron will be due to the magnetic field and will be acting as a centripetal force causing the electron to move in a circular path. (if the initial velocity of the electron had a component along the direction of the magnetic field it would have moved in a helical path)

Magnetic field(B)=6.5 G ( 1G = 10 ^{-4}T )

Speed of electron(v)=4.8 \times 10 ^ 6 ms ^{-1}

Charge of electron=-1.6\times 10^{-19}C

Mass of electron=9.1\times 10^{-31}kg

The angle between the direction of velocity and the magnetic field = 90o

Since the force due to the magnetic field is the only force acting on the particle,

\\\frac{mV^{2}}{r}=q\vec{V}\times \vec{B}\\ \frac{mV^{2}}{r}=|qVBsin\theta| \\ r=|\frac{mV}{qBsin\theta }|

r=\frac{9.1\times 10^{-31}\times 4.8\times 10^{6}}{1.6\times 10^{-19}\times 6.5\times 10^{-4}}=4.2 cm

 

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