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An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Fid the magnetic field and the number of revolutions per second made by the electron.

An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Fid the magnetic field and the number of revolutions per second made by the electron.

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
7 years ago
Sol. KE = 100 ev = 1.6 * 10^-17 J (1/2) * 9.1 * 10^-31 * V^2 = 1.6 * 10^-17 J ⇒ V^2 = 1.6 * 10^17 *2/9.1 * 10^-31 = 0.35 * 10^14 or, V = 0.591 * 10^7 m/s Now r = mυ/qB ⇒ 9.1 * 10^-31 *0.591 * 10^7/1.6 * 10^-19 * B = 10/100 ⇒ B = 9.1 * 0.591/1.6 * 10^-23/10^-19 = 3.3613 * 10^-4 T = 3.4 * 10^-4 T T = 2πm/qB = 2 * 3.14 * 9.1 * 10^-31/1.6 * 10^-19 * 3.4 * 10^-4 No. of Cycles per Second f = 1/T = 1.6 * 3.4/2 * 3.14 * 9.1 * 10^-19 * 10^-4/10^-31 = 0.0951 * 10^8 = 9.51 * 10^6 Note: ∴ Putting B ⃗ 3.361 * 10^-4 T We get f = 9.4 * 10^6

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