Question

The path of a moving charged particle with constant velocity entering a uniform magnetic field making an angle $\left(i\right)0^{\circ }\left(ii\right){45}^{\circ }\text{and}\left(iii\right){90}^{\circ }$ with the field will trace the path of-

• A. Circle ; helix ; parabola
• B. Straight line ; Circle ; ellipse
• C. Parabola ; straight line ; circle
• D. Straight line ; helix ; circle

Answer 1

The correct option is D Straight line ; helix ; circle

When, $\theta =0^{\circ };$

$\overrightarrow{v}\parallel \overrightarrow{B}\Rightarrow F=Bqv\sin \theta =0\to$ Straight line along the field.

When, $\theta ={45}^{\circ };$

$\overrightarrow{v}\parallel \overrightarrow{B}\Rightarrow$ Straight line motion

$\overrightarrow{v}\perp \overrightarrow{B}\Rightarrow$ uniform circular motion

Hence, the resultant motion will helical along the direction of field.

When, $\theta ={90}^{\circ };$

$\overrightarrow{v}\perp \overrightarrow{B}\Rightarrow$ uniform circular motion

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.