Sure, let's go through this step-by-step to find the displacement of a particle with charge [tex]\(q\)[/tex], mass [tex]\(m\)[/tex], and velocity [tex]\(\mathbf{v_0} = v_0(\hat{i} + \hat{j})\)[/tex] entering a region with a uniform magnetic field [tex]\(\mathbf{B_0} = B_0 \hat{i}\)[/tex].
### Step-by-Step Solution:
1. Magnetic Field and Particle Velocity:
- The magnetic field [tex]\(\mathbf{B_0} = B_0 \hat{i}\)[/tex] is aligned along the [tex]\(x\)[/tex]-axis.
- The particle's initial velocity is [tex]\(\mathbf{v_0} = v_0 (\hat{i} + \hat{j})\)[/tex], which has components along both the [tex]\(x\)[/tex]-axis and [tex]\(y\)[/tex]-axis.
2. Components of Velocity:
- The [tex]\(x\)[/tex]-component of velocity: [tex]\(v_{0x} = v_0\)[/tex]
- The [tex]\(y\)[/tex]-component of velocity: [tex]\(v_{0y} = v_0\)[/tex]
The velocity can be decomposed as:
[tex]\[
\mathbf{v_0} = v_0 \hat{i} + v_0 \hat{j}
\][/tex]
3. Motion in Magnetic Field:
- The magnetic force [tex]\( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \)[/tex].
- Given [tex]\( \mathbf{v} \)[/tex] with components in the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] directions and [tex]\( \mathbf{B} \)[/tex] in the [tex]\(x\)[/tex]-direction, the force in the [tex]\(z\)[/tex]-direction is affected.
- The [tex]\(x\)[/tex]-component of velocity, [tex]\(v_{0x}\)[/tex], is parallel to the magnetic field and does not experience any force due to the magnetic field.
- The [tex]\(y\)[/tex]-component of velocity, [tex]\(v_{0y}\)[/tex], and potential [tex]\(z\)[/tex]-velocity are perpendicular to the magnetic field, leading to circular motion in the [tex]\(yz\)[/tex]-plane.
4. Circular Motion:
- The radius of the circular path [tex]\(r = \frac{mv_{0y}}{qB}\)[/tex].
- The time period [tex]\(T = \frac{2\pi m}{q B}\)[/tex].
- After time [tex]\( t = \frac{\pi m}{q B} \)[/tex], the angle [tex]\(\theta\)[/tex] covered by the particle in its circular motion is [tex]\(\pi\)[/tex].
5. Displacement Calculation:
- The displacement due to circular motion in the [tex]\(yz\)[/tex]-plane after time [tex]\( t = \frac{\pi m}{q B} \)[/tex] when [tex]\(\theta = \pi\)[/tex].
- For circular motion, [tex]\(\Delta y = r\theta = \frac{mv_{0y}}{qB} \pi\)[/tex].
6. Net Displacement:
- Displacement along [tex]\(x\)[/tex]-axis remains constant because of no force in the [tex]\(x\)[/tex]-direction.
- Circular displacement in the [tex]\(yz\)[/tex]-plane but [tex]\(\hat{k}\)[/tex] remains similar.
- Total displacement: [tex]\( \frac{2mv_0}{qB}\)[/tex].
Given the choices:
1. [tex]\(\frac{m v_0}{q B}\left(\pi \hat{i} + 2 \hat{k}\right) \)[/tex]
2. [tex]\(\frac{m v_0}{q B}\left(\pi \hat{i}-2\hat{k}\right) \)[/tex]
3. [tex]\(\frac{2 \pi m v_0}{q B} \hat{i} \)[/tex]
4. [tex]\(\frac{2 \pi m v_0}{q B} \hat{j} \)[/tex]
The correct answer, based upon the evaluated process is:
(3) [tex]\(\frac{2 \pi m v_0}{q B} \hat{i}\)[/tex]
