JEE Physics 2024 Question with Solution

Given: A proton moves with constant velocity through a region of space.

Find: Which field configurations can make the net force on the proton zero.

For a charged particle, the Lorentz force is

$$\vec{F} = q\left(\vec{E} + \vec{v} \times \vec{B}\right)$$

Since the velocity does not change, the net force must be zero:

$$\vec{E} + \vec{v} \times \vec{B} = 0$$

Now examine each case.

1. Option A: $\vec{E} = 0, \vec{B} = 0$

Then

$$\vec{F} = 0$$

So constant velocity is possible.

2. Option B: $\vec{E} = 0, \vec{B} \ne 0$

Then the force is

$$\vec{F} = q\left(\vec{v} \times \vec{B}\right)$$

This can still be zero if $\vec{v}$ is parallel to $\vec{B}$. So this case is possible.

3. Option C: $\vec{E} \ne 0, \vec{B} = 0$

Then

$$\vec{F} = q\vec{E}$$

which is non-zero because $q \ne 0$ and $\vec{E} \ne 0$. So constant velocity is not possible.

4. Option D: $\vec{E} \ne 0, \vec{B} \ne 0$

This is possible if the electric and magnetic forces cancel:

$$\vec{E} = -\left(\vec{v} \times \vec{B}\right)$$

Hence this case is also possible.

Therefore, the correct options are A, B and D.

The solution concludes these cases are correct, although the solution incorrectly states C. The working supports A, B and D.