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:
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:
Based on this analysis, the options where the proton can move with constant velocity are (A), (B), and (D).
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).
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals: