A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $$2 \times 10^{5} ms^{-1}$$. When the electric field is switched off, the proton moves along a circular path of radius 2 cm .The magnitude of electric field is $$x\times 10^{4} N/C$$. The value of x is________Take the mass of the proton $$1.6\times 10^{-27}kg$$

We are told a proton moves undeflected in crossed electric and magnetic fields at $$v = 2 \times 10^5$$ m/s. When the electric field is switched off, the proton moves in a circle of radius $$r = 2$$ cm. We need to find the electric field.

Condition for undeflected motion in crossed fields

When a charged particle moves undeflected through crossed electric and magnetic fields, the electric force and magnetic force balance each other:

$$qE = qvB$$

$$E = vB \quad \ldots (1)$$

Find $$B$$ from the circular motion when $$E$$ is switched off

When the electric field is switched off, only the magnetic force acts, providing the centripetal force for circular motion:

$$qvB = \frac{mv^2}{r}$$

$$B = \frac{mv}{qr} \quad \ldots (2)$$

Substituting given values ($$m = 1.6 \times 10^{-27}$$ kg, $$v = 2 \times 10^5$$ m/s, $$q = 1.6 \times 10^{-19}$$ C, $$r = 2$$ cm $$= 0.02$$ m):

$$B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02}$$

$$B = \frac{3.2 \times 10^{-22}}{3.2 \times 10^{-21}} = \frac{1}{10} = 0.1 \text{ T}$$

Calculate the electric field using equation (1)

$$E = vB = 2 \times 10^5 \times 0.1 = 2 \times 10^4 \text{ N/C}$$

Therefore, $$x = 2$$.

The answer is 2.