An electron with an energy of 4.8 · 10-15 J flies into a uniform magnetic field with an induction of 0.01

| April 28, 2021 | education

An electron with an energy of 4.8 · 10-15 J flies into a uniform magnetic field with an induction of 0.01 T perpendicular to the induction lines. What is the radius of curvature of the trajectory of the electron?

To find the value of the radius of curvature of the trajectory of a given electron, we will use the equality: Fl (Lorentz force) = qe * V * B = me * a = me * V ^ 2 / R, whence we can express: R = me * V ^ 2 / (qe * V * B) = me * V / (qe * B) = me * √ (2Ek / me) / (qe * B).

Data: me – electron mass (me = 9.11 * 10 ^ -31 kg); Ek is the kinetic energy of a given electron (Ek = 4.8 * 10 ^ -15 J); qe – electron charge (qe = -1.6 * 10 ^ -19 C); B – induction (B = 0.01 T).

Let’s calculate: R = me * √ (2Ek / me) / (qe * B) = 9.11 * 10 ^ -31 * √ (2 * 4.8 * 10 ^ -15 / 9.11 * 10 ^ -31 ) / (1.6 * 10 ^ -19 * 0.01) = 0.058 m or 58 mm.

Answer: The radius of curvature of the trajectory of a given electron is 58 mm.



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