Question

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If \( \vec{E} \) and \( \vec{B} \) represent the electric and magnetic fields respectively, then the region of space may have:
(A) \( \vec{E} = 0, \vec{B} = 0 \) 
(B) \( \vec{E} = 0, \vec{B} \neq 0 \) 
(C) \( \vec{E} \neq 0, \vec{B} = 0 \) 
(D) \( \vec{E} \neq 0, \vec{B} \neq 0 \) 
Choose the most appropriate answer from the options given below:

• A. (A), (B) and (C) only
• B. (A), (C) and (D) only
• C. (A), (B) and (D) only
• D. (B), (C) and (D) only

Answer 1

The Correct Option is C

This problem requires analyzing the interactions of a proton moving at a constant velocity through electric and magnetic fields. Constant velocity implies a net force of zero on the proton.

The Lorentz force law describes the forces on a charged particle in electric and magnetic fields:

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

where:

• \(F\) is the total force on the particle,
• \(q\) is the particle's charge (positive for a proton),
• \(\vec{E}\) is the electric field,
• \(\vec{v}\) is the particle's velocity,
• \(\vec{B}\) is the magnetic field.

For constant velocity, the net force \(F\) must be zero:

\(\vec{E} + \vec{v} \times \vec{B} = 0\)

We will now evaluate each scenario:

1. Case (A): \( \vec{E} = 0, \vec{B} = 0 \)
   • With both fields absent, there is no force, allowing constant velocity. Thus, (A) is valid.
2. Case (B): \( \vec{E} = 0, \vec{B} eq 0 \)
   • If the electric field is zero, the cross product term \( \vec{v} \times \vec{B} \) must also be zero for zero net force.
   • This occurs if \( \vec{v} \) is parallel to \( \vec{B} \), or if \( \vec{B} \) is zero.
   • Therefore, (B) is also valid.
3. Case (C): \( \vec{E} eq 0, \vec{B} = 0 \)
   • With \( \vec{B} = 0 \), only the electric field exerts a force.
   • An electric field will alter the velocity unless \( \vec{E} = 0 \).
   • Consequently, (C) is invalid.
4. Case (D): \( \vec{E} eq 0, \vec{B} eq 0 \)
   • Zero net force requires the electric and magnetic forces to cancel: \( \vec{E} = -\vec{v} \times \vec{B} \).
   • This is achievable with specific \( \vec{E}, \vec{v}, \) and \( \vec{B} \) orientations.
   • Hence, (D) can be valid.

Based on this analysis, the proton can move with constant velocity in cases (A), (B), and (D).