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प्रश्न
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.
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उत्तर १
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
उत्तर २
When a charged particle enters a magnetic field with its velocity making an angle with the field, the motion splits into two components:
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A component parallel to the magnetic field: motion is uniform linear motion.
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A component perpendicular to the magnetic field: motion is circular due to the magnetic force (Lorentz force).
Since both motions happen simultaneously:
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The perpendicular component causes circular motion, and
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The parallel component causes the particle to move forward along the magnetic field.
v⊥ = v sinθ = 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
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Path shape: Helix
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Diameter: 0.36 m
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Pitch: 0.55 m
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