Question

A particle with a charge of 5.0 µC and a mass of 5.0 × 10⁻¹² kg is projected with a speed of 1.0 km s⁻¹ in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin⁻¹ (0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

Answer 1

Given:
Charge of the particle, q = 5 µC = 5 × 10⁻⁶ C
Magnetic field intensity, B = 5 × 10⁻³ T
Mass of the particle, m = 5 × 10⁻¹² kg
Velocity of projection, v = 1 Km/s = 10³ m/s
Angle between the magnetic field and velocity, θ= sin⁻¹(0.9)
Component of velocity perpendicular to the magnetic field, `v _⊥= v sin theta`
Component of velocity in the direction of magnetic field, `v_(||)` 
Since there are no forces in the horizontal direction (the direction of magnetic field),  the particle moves with uniform velocity.
The velocity has a vertical component along which it accelerates with an acceleration a and moves in a circular cross-section. Thus, it moves in a helix.
`(mv_⊥ B)`
⇒ r = `"mvsinθ "/(qB)`
      = `(5xx10^-12xx10^3xx0.90)/(5xx10^-6xx 5 xx10^3)`
Hence, diameter of the helix, 2r = 0.36 m = 36 cm
Pitch,
`P = (2pir)/"vsinθ"xx sqrt(1 - 0.81`
= 0.55 m = 55cm

Answer 2

When a charged particle enters a magnetic field with its velocity making an angle with the field, the motion splits into two components:

• A component parallel to the magnetic field: motion is uniform linear motion.

• A component perpendicular to the magnetic field: motion is circular due to the magnetic force (Lorentz force).

Since both motions happen simultaneously:

• The perpendicular component causes circular motion, and

• The parallel component causes the particle to move forward along the magnetic field.

v⊥ = v sin⁡θ = 1.0 × 10³ × 0.90 = 900 m/s 

`v|| = v costheta = 1.0 xx10^3 xx sqrt(1-0.90^2) = 1.0xx10^3 xx sqrt0.19 = 436.8` m/s

The magnetic force provides centripetal acceleration:

`r = (mv⊥)/(qB) = (5.0xx10^-12 xx900)/(5.0 xx 10^-6xx5.0 xx10^-3) = (4.5xx10^-9)/(2.5xx10^-8) = 0.18 m`

Diameter D = 2r = 2 × 0.18 = 0.36 m

Time for one circle (T) = `(2pim)/(qB) = (2pixx5.0 xx10^-12)/(5.0 xx 10^-6xx5.0xx10^-3) = (10pi xx 10^-12)/(2.5xx 10^-8) = 1.257 xx 10^-3`

Pitch = v∥ ​× T = 436.8 × 1.257 × 10⁻³ ≈ 0.55m​

• Path shape: Helix

• Diameter: 0.36 m

• Pitch: 0.55 m