0

Guest

Courses New

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
9 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

Think You Can Provide A Better Answer ?

Other Related Questions on Electromagnetic Induction

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More