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`        A proton goes undeflected in a crossed in a crossed electric and magnetic field (the field are perpendicular to each other) at a speed of 2.0 × 10^5 m/s. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton = 1.6 × 10^-27 kg. `
3 years ago

396 Points
```										Sol. M base P = 1.6 × 10^–27 Kg
υ = 2 × 105 m/s r = 4 cm = 4 × 10^–2 m
Since the proton is undeflected in the combined magnetic and electric field. Hence force due to both the fields must be same.
i.e. qE = qυB ⇒ E = υB
Won, when the electricfield is stopped, then if forms a circle due to force of magnetic field
We know r = mυ/qB
⇒ 4 * 10^2 = 1.6 *10^-27 *2 * 10^5/1.6 * 10^-19 *B
⇒ B = 1.6 *10^-27 *2 * 10^5/4 * 10^2 * 1.6 *10^-19 = 0.5 * 10^-1 = 0.005 T
E = υB = 2 * 10^5 * 0.05 = 1 * 10^4 N/C

```
3 years ago
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