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Magnetics question

2020 · 2 Sep · Shift 1 · Q59
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Magnetics question

2020 · 2 Sep · Shift 1 · Q59

JEE MainPhysicsMagneticsMCQ+4 / −1
A beam of protons with speed 4 × 105 ms–1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton = 1.67 ×\times× 10–27 kg, charge of the proton = 1.69 ×\times× 10–19 C)
  1. A
    2 cm
  2. B
    12 cm
  3. C
    5 cm
  4. D
    4 cm
View written solutionFree

Correct answer: D

  1. Given data
  • Speed of proton: v=4×105 m s−1v = 4 \times 10^5\ \text{m s}^{-1}v=4×105 m s−1
  • Magnetic field: B=0.3 TB = 0.3\ \text{T}B=0.3 T
  • Angle with magnetic field: θ=60∘\theta = 60^\circθ=60∘
  • Mass of proton: m=1.67×10−27 kgm = 1.67 \times 10^{-27}\ \text{kg}m=1.67×10−27 kg
  • Charge of proton: q=1.69×10−19 Cq = 1.69 \times 10^{-19}\ \text{C}q=1.69×10−19 C
  1. Concept of pitch of helical path

When a charged particle enters a magnetic field at an angle, its velocity has two components:

  • Parallel to field: v∥=vcos⁡θv_\parallel = v \cos\thetav∥​=vcosθ
  • Perpendicular to field: v⊥=vsin⁡θv_\perp = v \sin\thetav⊥​=vsinθ

The perpendicular component causes circular motion, while the parallel component causes forward motion along the field. Hence the path is helical.

The pitch is the distance moved along the field in one time period:

p=v∥Tp = v_\parallel Tp=v∥​T

where the time period is

T=2πmqBT = \frac{2\pi m}{qB}T=qB2πm​

So,

p=vcos⁡θ⋅2πmqBp = v\cos\theta \cdot \frac{2\pi m}{qB}p=vcosθ⋅qB2πm​
  1. Calculate the parallel component
v∥=vcos⁡60∘=4×105×12=2×105 m s−1v_\parallel = v\cos 60^\circ = 4 \times 10^5 \times \frac{1}{2} = 2 \times 10^5\ \text{m s}^{-1}v∥​=vcos60∘=4×105×21​=2×105 m s−1
  1. Calculate the time period
T=2πmqB=2π×1.67×10−271.69×10−19×0.3T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.69 \times 10^{-19} \times 0.3}T=qB2πm​=1.69×10−19×0.32π×1.67×10−27​

First compute denominator:

1.69×0.3=0.5071.69 \times 0.3 = 0.5071.69×0.3=0.507

So,

T=2π×1.67×10−270.507×10−19T = \frac{2\pi \times 1.67 \times 10^{-27}}{0.507 \times 10^{-19}}T=0.507×10−192π×1.67×10−27​ T=2π×1.670.507×10−8T = 2\pi \times \frac{1.67}{0.507} \times 10^{-8}T=2π×0.5071.67​×10−8 1.670.507≈3.29\frac{1.67}{0.507} \approx 3.290.5071.67​≈3.29

Thus,

T≈2π×3.29×10−8T \approx 2\pi \times 3.29 \times 10^{-8}T≈2π×3.29×10−8 T≈20.67×10−8=2.067×10−7 sT \approx 20.67 \times 10^{-8} = 2.067 \times 10^{-7}\ \text{s}T≈20.67×10−8=2.067×10−7 s
  1. Calculate pitch
p=v∥T=2×105×2.067×10−7p = v_\parallel T = 2 \times 10^5 \times 2.067 \times 10^{-7}p=v∥​T=2×105×2.067×10−7 p=4.134×10−2 mp = 4.134 \times 10^{-2}\ \text{m}p=4.134×10−2 m p=0.04134 m=4.13 cmp = 0.04134\ \text{m} = 4.13\ \text{cm}p=0.04134 m=4.13 cm
  1. Closest option

The pitch is closest to

4 cm\boxed{4\ \text{cm}}4 cm​

So the correct option is D.

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