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 $=1.6 \times 10^{-27} \mathrm{~kg}$.

<p>Let's break down the problem step by step:</p> <p><p>When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,</p> <p>$qE = qvB,$</p> <p>which simplifies to</p> <p>$E = vB.$</p></p> <p><p>After the electric field is switched off, the proton moves in a circular path under the action of the magnetic force. The magnetic force provides the required centripetal force:</p> <p>$qvB = \frac{mv^2}{r}.$</p> <p>Solving for the magnetic field $B$, we get:</p> <p>$B = \frac{mv}{qr}.$</p></p> <p><p>Now substitute this expression for $B$ back into the equilibrium condition:</p> <p>$E = vB = v\left(\frac{mv}{qr}\right) = \frac{mv^2}{qr}.$</p></p> <p><p>Plug in the given values:</p></p> <p><p>Proton mass, $m = 1.6 \times 10^{-27} \, \text{kg}$</p></p> <p><p>Speed, $v = 2 \times 10^5 \, \text{m/s}$</p></p> <p><p>Radius, $r = 2 \, \text{cm} = 0.02 \, \text{m}$</p></p> <p><p>Proton charge, $q = 1.6 \times 10^{-19} \, \text{C}$</p> <p>Thus,</p> <p>$$E = \frac{(1.6 \times 10^{-27} \, \text{kg})(2 \times 10^5 \, \text{m/s})^2}{(1.6 \times 10^{-19} \, \text{C})(0.02 \, \text{m})}.$$</p></p> <p>Calculate the numerator:</p> <p><p>$(2 \times 10^5)^2 = 4 \times 10^{10},$</p></p> <p><p>So, $(1.6 \times 10^{-27}) \times (4 \times 10^{10}) = 6.4 \times 10^{-17}.$</p></p> <p><p>Calculate the denominator:</p> <p>$(1.6 \times 10^{-19}) \times (0.02) = 3.2 \times 10^{-21}.$</p></p> <p><p>Now compute the electric field:</p> <p>$$E = \frac{6.4 \times 10^{-17}}{3.2 \times 10^{-21}} = 2 \times 10^4 \, \text{N/C}.$$</p></p> <p>The problem states that the magnitude of the electric field is $x \times 10^4 \, \text{N/C}.$ Since we found</p> <p>$E = 2 \times 10^4 \, \text{N/C},$</p> <p>it follows that</p> <p>$x = 2.$</p>