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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.`
3 years ago

Kevin Nash
332 Points
```										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

```
3 years ago
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