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Home » In a chamber, a uniform magnetic field of 6.5 G (1 G = T) is maintained. An electron is shot into the field with a speed of normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
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 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | 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 \times {10^6}m/s normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

Magnetic field strength (B) = 6.5 G = 6.5 \times {10^{ - 4}}T

Speed of the electron (v) = 4.8 \times {10^6}m/s

Charge on the electron (e) = 1.6 \times {10^{ - 19}}C

Mass of the electron (me) = 9.1 \times {10^{ - 31}}kg

Angle formed by the fired electron and the magnetic field, \theta = {90^ \circ }

The magnetic force exerted on the electron in a magnetic field is given by the relationship:F = evB\sin \theta

The travelling electron receives centripetal force from this force. As a result, the electron begins to move in a circular route of radius r.

Hence, centripetal force exerted on the electron,

{F_e} = \frac{{m{v^2}}}{r}

The centripetal force imparted on the electron is equal to the magnetic force at equilibrium, i.e.,

Fe = F

\frac{{m{v^2}}}{r} = evB\sin \theta

r = \frac{{mv}}{{eB\sin \theta }}

So,

r = \frac{{9.1 \times {{10}^{ - 31}} \times 4.8 \times {{10}^6}}}{{6.5 \times {{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}} \times \sin {{90}^ \circ }}}

r = 4.2cm

As a result, the radius of the electron’s circular orbit is 4.2 cm.

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