QJEEM25Ph558f: JEE Main 2025 Physics - Class 12 JEEM25Ph558f Solution: A proton is moving undeflected in a region of crossed electric…

Let's break down the problem step by step:

When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,

$qE = qvB,$

which simplifies to

$E = vB.$

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:

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

Solving for the magnetic field $B$ , we get:

$B = \frac{mv}{qr}.$

Now substitute this expression for $B$ back into the equilibrium condition:

$E = vB = v\left(\frac{mv}{qr}\right) = \frac{mv^2}{qr}.$

Plug in the given values:

Proton mass, $m = 1.6 \times 10^{-27} \, \text{kg}$

Speed, $v = 2 \times 10^5 \, \text{m/s}$

Radius, $r = 2 \, \text{cm} = 0.02 \, \text{m}$

Proton charge, $q = 1.6 \times 10^{-19} \, \text{C}$

Thus,

$E = \frac{(1.6 \times 10^{-27} \, \text{kg})(2 \times 10^5 \, \text{m/s})^2}{(1.6 \times 10^{-19} \, \text{C})(0.02 \, \text{m})}.$

Calculate the numerator:

$(2 \times 10^5)^2 = 4 \times 10^{10},$

So, $(1.6 \times 10^{-27}) \times (4 \times 10^{10}) = 6.4 \times 10^{-17}.$

Calculate the denominator:

$(1.6 \times 10^{-19}) \times (0.02) = 3.2 \times 10^{-21}.$

Now compute the electric field:

$E = \frac{6.4 \times 10^{-17}}{3.2 \times 10^{-21}} = 2 \times 10^4 \, \text{N/C}.$

The problem states that the magnitude of the electric field is $x \times 10^4 \, \text{N/C}.$ Since we found

$E = 2 \times 10^4 \, \text{N/C},$

it follows that

$x = 2.$