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Magnetics question

2023 · 6 Apr · Shift 2 · Q69
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  5. /2023 · 6 Apr · Shift 2 · Q69

Magnetics question

2023 · 6 Apr · Shift 2 · Q69

JEE MainPhysicsMagneticsNumerical+4 / −1
A proton with a kinetic energy of 2.0 eV2.0 ~\mathrm{eV}2.0 eV moves into a region of uniform magnetic field of magnitude π2×10−3 T\frac{\pi}{2} \times 10^{-3} \mathrm{~T}2π​×10−3 T. The angle between the direction of magnetic field and velocity of proton is 60∘60^{\circ}60∘. The pitch of the helical path taken by the proton is ‾cm\underline{\hspace{2cm}}\mathrm{cm}​cm. (Take, mass of proton =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg}=1.6×10−27 kg and Charge on proton =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}=1.6×10−19C ).
Numerical answer
View written solutionFree

Correct answer: 40

  1. Given data
  • Kinetic energy of proton: K=2.0 eVK = 2.0\,\text{eV}K=2.0eV
  • Magnetic field: B=π2×10−3 TB = \dfrac{\pi}{2}\times 10^{-3}\,\text{T}B=2π​×10−3T
  • Angle between v⃗\vec vv and B⃗\vec BB: 60∘60^\circ60∘
  • Mass of proton: m=1.6×10−27 kgm = 1.6\times 10^{-27}\,\text{kg}m=1.6×10−27kg
  • Charge of proton: q=1.6×10−19 Cq = 1.6\times 10^{-19}\,\text{C}q=1.6×10−19C

We need the pitch of the helical path.

  1. Find the speed of the proton

Convert kinetic energy into joules:

K=2.0 eV=2.0×1.6×10−19=3.2×10−19 JK = 2.0\,\text{eV} = 2.0\times 1.6\times 10^{-19} = 3.2\times 10^{-19}\,\text{J}K=2.0eV=2.0×1.6×10−19=3.2×10−19J

Using

K=12mv2K = \frac{1}{2}mv^2K=21​mv2

so,

v=2Km=2×3.2×10−191.6×10−27v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2\times 3.2\times 10^{-19}}{1.6\times 10^{-27}}}v=m2K​​=1.6×10−272×3.2×10−19​​ v=4×108=2×104 m/sv = \sqrt{4\times 10^8} = 2\times 10^4\,\text{m/s}v=4×108​=2×104m/s
  1. Resolve velocity along magnetic field

The component parallel to the magnetic field is:

v∥=vcos⁡60∘=2×104×12=104 m/sv_\parallel = v\cos 60^\circ = 2\times 10^4 \times \frac{1}{2} = 10^4\,\text{m/s}v∥​=vcos60∘=2×104×21​=104m/s

This component remains unchanged and is responsible for the pitch.

  1. Find the time period of circular motion

For a charged particle in a magnetic field,

T=2πmqBT = \frac{2\pi m}{qB}T=qB2πm​

Substitute B=π2×10−3B = \dfrac{\pi}{2}\times 10^{-3}B=2π​×10−3:

T=2π×1.6×10−271.6×10−19×(π2×10−3)T = \frac{2\pi \times 1.6\times 10^{-27}}{1.6\times 10^{-19}\times \left(\frac{\pi}{2}\times 10^{-3}\right)}T=1.6×10−19×(2π​×10−3)2π×1.6×10−27​

Cancel 1.61.61.6 and π\piπ:

T=2×10−2710−19×(12×10−3)T = \frac{2\times 10^{-27}}{10^{-19}\times \left(\frac{1}{2}\times 10^{-3}\right)}T=10−19×(21​×10−3)2×10−27​

More directly,

T=2πmqB=2πmq(π2×10−3)=4mq×10−3T = \frac{2\pi m}{qB} = \frac{2\pi m}{q\left(\frac{\pi}{2}\times 10^{-3}\right)} = \frac{4m}{q\times 10^{-3}}T=qB2πm​=q(2π​×10−3)2πm​=q×10−34m​ T=4×1.6×10−271.6×10−19×10−3=4×10−5 sT = \frac{4\times 1.6\times 10^{-27}}{1.6\times 10^{-19}\times 10^{-3}} = 4\times 10^{-5}\,\text{s}T=1.6×10−19×10−34×1.6×10−27​=4×10−5s
  1. Calculate the pitch

Pitch ppp is the distance moved along the field in one time period:

p=v∥Tp = v_\parallel Tp=v∥​T p=104×4×10−5=4×10−1 mp = 10^4 \times 4\times 10^{-5} = 4\times 10^{-1}\,\text{m}p=104×4×10−5=4×10−1m p=0.4 m=40 cmp = 0.4\,\text{m} = 40\,\text{cm}p=0.4m=40cm
  1. Final answer
40\boxed{40}40​

The pitch of the helical path is 40 cm40\,\text{cm}40cm.

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