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

For a proton to move with a constant velocity without any change, the net force on the particle must be zero. This implies:  
\( q\vec{E} + q\vec{v} \times \vec{B} = 0 \)

Possible cases that satisfy this condition are:

Step 1. Case (A): \( \vec{E} = 0 \) and \( \vec{B} = 0 \) — No electric or magnetic fields are present, so no force acts on the proton.

Step 2. Case (B): \( \vec{E} = 0 \) and \( \vec{B} \neq 0 \) — The proton experiences no electric force, and if \( \vec{v} \) and \( \vec{B} \) are parallel, \( \vec{v} \times \vec{B} = 0 \).

Step 3. Case (D): \( \vec{E} \neq 0 \) and \( \vec{B} \neq 0 \) — Here, \( q\vec{E} \) and \( q\vec{v} \times \vec{B} \) can cancel each other out if they are equal in magnitude and opposite in direction.

Thus, the region of space may satisfy cases (A), (B), and (D), so the correct answer is (3).