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

Step 1: Understand Given Information

• Speed of proton: $v = 2 \times 10^5$ m/s
• Radius of circular path when $E = 0$: $r = 2$ cm = $0.02$ m
• Mass of proton: $m = 1.6 \times 10^{-27}$ kg
• Charge of proton: $q = 1.6 \times 10^{-19}$ C

Step 2: Find the Magnetic Field $B$

When the electric field is switched off, the proton moves in a circular path due to the magnetic force. The centripetal force is provided by the magnetic force:

$$q v B = \frac{m v^2}{r}$$

Rearrange for $B$:

$$B = \frac{m v}{q r}$$

Substituting values:

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

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

Step 3: Find the Electric Field $E$

Since the proton is moving undeflected in the region of crossed $E$ and $B$ fields, the electric force and magnetic force must balance each other:

$$q E = q v B$$

$$E = v B$$

Substituting values:

$$E = (2 \times 10^5) (0.1)$$

 $$E = 2 \times 10^4 \text{ N/C}$$ 

Thus, $x = \mathbf{2}$.