JEE Physics 2025 Question with Solution

Given: The proton moves undeflected with speed $2 \times 10^5 \, \text{m/s}$ in crossed electric and magnetic fields. When the electric field is removed, it moves in a circle of radius $0.02 \, \text{m}$. Mass of proton is $1.6 \times 10^{-27} \, \text{kg}$ and charge of proton is $1.6 \times 10^{-19} \, \text{C}$.

Find: The value of $x$ in $E = x \times 10^4 \, \text{N/C}$.

For undeflected motion in crossed fields, electric force balances magnetic force:

$$qE = qvB$$

So,

$$E = vB$$

When the electric field is switched off, magnetic force provides the centripetal force:

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

Hence,

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

Substituting the given values:

$$B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02}$$ $$B = 0.01 \, \text{T}$$

Now,

$$E = vB = 2 \times 10^5 \times 0.01 = 2000 \, \text{N/C}$$

So,

$$x \times 10^4 = 2000$$ $$x = 0.2$$

Therefore, the value of $x$ is $0.2$.

The solution also shows a conflicting listed answer of $1$, but the worked steps lead to $0.2$, so $0.2$ is the correct value.

Given: The proton is undeflected in crossed fields and then moves in a circular path when only magnetic field acts.

Find: The value of $x$.

Use the two direct relations:

$$E = vB$$

and

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

Combine them:

$$E = v \cdot \frac{mv}{qr} = \frac{mv^2}{qr}$$

Substitute values:

$$E = \frac{1.6 \times 10^{-27} \times (2 \times 10^5)^2}{1.6 \times 10^{-19} \times 0.02}$$ $$E = 2000 \, \text{N/C} = 0.2 \times 10^4 \, \text{N/C}$$

Therefore, $x = 0.2$.