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) \( \vec{E} = 0, \vec{B} = 0 \)
(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), (B) and (C) only
- (A), (C) and (D) only
- (A), (B) and (D) only
- (B), (C) and (D) only
The Correct Option is C
Approach Solution - 1
In this problem, we need to analyze the interactions of a proton moving through a region of space with electric and magnetic fields. We are given that the proton moves with constant velocity, indicating that the net force on it is zero.
The forces experienced by a charged particle (like a proton) moving in electric and magnetic fields are given by the Lorentz force law:
\(F = q(\vec{E} + \vec{v} \times \vec{B})\)
where:
- \(F\) is the total force on the particle,
- \(q\) is the charge of the particle (positive for a proton),
- \(\vec{E}\) is the electric field,
- \(\vec{v}\) is the velocity of the particle,
- \(\vec{B}\) is the magnetic field.
For the proton to continue moving with constant velocity, the net force \(F\) must be zero:
\(\vec{E} + \vec{v} \times \vec{B} = 0\)
We will now analyze each given scenario:
- Case (A): \(\vec{E} = 0, \vec{B} = 0\)
- Here, both fields are zero, so there is no force, allowing the proton to move with constant velocity. This is possible, so (A) is correct.
- Case (B): \(\vec{E} = 0, \vec{B} \neq 0\)
- If the electric field is zero, to have no net force, the cross product term \(\vec{v} \times \vec{B}\) must also be zero.
- This is possible if \(\vec{v}\) is parallel to \(\vec{B}\) or \(\vec{B}\) is zero itself.
- Therefore, (B) is also correct.
- Case (C): \(\vec{E} \neq 0, \vec{B} = 0\)
- With \(\vec{B} = 0\), the magnetic force is zero, and only the electric field contributes to the force.
- The presence of an electric field will exert a force, disturbing the constant velocity condition unless \(\vec{E} = 0\).
- So, (C) is not correct.
- Case (D): \(\vec{E} \neq 0, \vec{B} \neq 0\)
- To achieve zero net force, the electric and magnetic forces must cancel each other, i.e., \(\vec{E} = -\vec{v} \times \vec{B}\).
- This is possible with specific orientations of \(\vec{E}, \vec{v},\) and \(\vec{B}\).
- Thus, (D) can be correct.
Based on this analysis, the options where the proton can move with constant velocity are (A), (B), and (D).
Approach Solution -2
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).
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