Moving charges in magnetic field — force, circular motion, helical path

Question

A proton enters a uniform magnetic field $B = 0.5$ T with velocity $v = 10^6$ m/s at an angle of $30°$ to the field. Find the radius of the circular component and the pitch of the helical path. ( $m_p = 1.67 \times 10^{-27}$ kg, $q = 1.6 \times 10^{-19}$ C)

(JEE Main 2024 pattern)

Solution — Step by Step

When the velocity is at angle $\theta$ to $\vec{B}$ :

$v_\perp = v\sin\theta = 10^6 \times \sin 30° = 5 \times 10^5$ m/s (creates circular motion)
$v_\parallel = v\cos\theta = 10^6 \times \cos 30° = 5\sqrt{3} \times 10^5$ m/s (unchanged, moves along field)

The parallel component is unaffected by the magnetic force ( $\vec{F} = q\vec{v} \times \vec{B}$ is zero when $\vec{v} \parallel \vec{B}$ ).

The perpendicular component creates circular motion. Equating magnetic force to centripetal force:

qv_\perp B = \frac{mv_\perp^2}{r}

r = \frac{mv_\perp}{qB} = \frac{1.67 \times 10^{-27} \times 5 \times 10^5}{1.6 \times 10^{-19} \times 0.5} = \mathbf{0.0104 \text{ m} \approx 1.04 \text{ cm}}

Time period of circular motion (independent of speed):

T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.5} = 1.31 \times 10^{-7} \text{ s}

Pitch = distance moved along the field in one revolution:

p = v_\parallel \times T = 5\sqrt{3} \times 10^5 \times 1.31 \times 10^{-7} \approx \mathbf{0.113 \text{ m}}

Why This Works

The magnetic force is always perpendicular to velocity — it changes direction but not speed. The key insight is the angle between $\vec{v}$ and $\vec{B}$ :

graph TD
    A["Charge enters B field"] --> B{"Angle between v and B?"}
    B -->|"θ = 0° (parallel)"| C["No force<br/>Straight line path"]
    B -->|"θ = 90° (perpendicular)"| D["Circular path<br/>r = mv/qB"]
    B -->|"0° < θ < 90°"| E["Helical path"]
    E --> F["Circular component: v⊥ = v sinθ"]
    E --> G["Linear component: v∥ = v cosθ"]
    F --> H["Radius = mv⊥/qB"]
    G --> I["Pitch = v∥ × T"]
    D --> J["Time period T = 2πm/qB<br/>(independent of v!)"]

Alternative Method — Using the Lorentz Force Directly

Write $\vec{F} = q(\vec{v} \times \vec{B})$ in component form. If $\vec{B} = B\hat{z}$ and $\vec{v} = v_x\hat{x} + v_z\hat{z}$ , then $\vec{F} = qv_x B\hat{y}$ — force only in the plane perpendicular to $B$ . This confirms circular motion in the xy-plane while $v_z$ stays constant.

The time period $T = 2\pi m/(qB)$ does not depend on velocity. This is why cyclotrons work — particles of all speeds complete one revolution in the same time, staying in sync with the alternating voltage. JEE loves asking “what remains constant: radius, time period, or frequency?” Answer: time period and frequency.

Common Mistake

Students use the total velocity $v$ instead of $v_\perp$ in the radius formula when the charge enters at an angle. The radius depends only on the perpendicular component: $r = mv_\perp/(qB)$ . Using total $v$ gives a larger radius and wrong pitch. Always decompose first.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Lorentz Force Directly

Common Mistake

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