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Magnetics question

2019 · 11 Jan · Shift 1 · Q64
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Magnetics question

2019 · 11 Jan · Shift 1 · Q64

JEE MainPhysicsMagneticsMCQ+4 / −1
In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 ×\times× 10–19 C Mass of the electron = 9.1 ×\times× 10–31 kg]
  1. A
    7.5 ×\times× 10 −-− 4 m
  2. B
    7.5 ×\times× 10 −-− 3 m
  3. C
    7.5 m
  4. D
    7.5 ×\times× 10 −-− 2 m
View written solutionFree

Correct answer: A

  1. Find the speed of the electron after acceleration

When an electron is accelerated from rest through a potential difference VVV, the gained kinetic energy is:

eV=12mv2eV = \frac{1}{2}mv^2eV=21​mv2

Given:

  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C
  • V=500 VV = 500\,\text{V}V=500V
  • m=9.1×10−31 kgm = 9.1 \times 10^{-31}\,\text{kg}m=9.1×10−31kg

So,

1.6×10−19×500=12(9.1×10−31)v21.6 \times 10^{-19} \times 500 = \frac{1}{2}(9.1 \times 10^{-31})v^21.6×10−19×500=21​(9.1×10−31)v2

8.0×10−17=12(9.1×10−31)v28.0 \times 10^{-17} = \frac{1}{2}(9.1 \times 10^{-31})v^28.0×10−17=21​(9.1×10−31)v2

v2=2×8.0×10−179.1×10−31v^2 = \frac{2 \times 8.0 \times 10^{-17}}{9.1 \times 10^{-31}}v2=9.1×10−312×8.0×10−17​

v2=1.6×10−169.1×10−31≈1.758×1014v^2 = \frac{1.6 \times 10^{-16}}{9.1 \times 10^{-31}} \approx 1.758 \times 10^{14}v2=9.1×10−311.6×10−16​≈1.758×1014

v≈1.33×107 m/sv \approx 1.33 \times 10^7\,\text{m/s}v≈1.33×107m/s

  1. Use magnetic force as centripetal force

When the electron enters a magnetic field perpendicular to its velocity, it moves in a circular path of radius rrr:

evB=mv2revB = \frac{mv^2}{r}evB=rmv2​

So,

r=mveBr = \frac{mv}{eB}r=eBmv​

Given:

  • m=9.1×10−31 kgm = 9.1 \times 10^{-31}\,\text{kg}m=9.1×10−31kg
  • v=1.33×107 m/sv = 1.33 \times 10^7\,\text{m/s}v=1.33×107m/s
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C
  • B=100 mT=0.1 TB = 100\,\text{mT} = 0.1\,\text{T}B=100mT=0.1T

Substitute:

r=9.1×10−31×1.33×1071.6×10−19×0.1r = \frac{9.1 \times 10^{-31} \times 1.33 \times 10^7}{1.6 \times 10^{-19} \times 0.1}r=1.6×10−19×0.19.1×10−31×1.33×107​

r=1.2103×10−231.6×10−20r = \frac{1.2103 \times 10^{-23}}{1.6 \times 10^{-20}}r=1.6×10−201.2103×10−23​

r≈0.756×10−3r \approx 0.756 \times 10^{-3}r≈0.756×10−3

r≈7.56×10−4 mr \approx 7.56 \times 10^{-4}\,\text{m}r≈7.56×10−4m

  1. Match with the options

r≈7.5×10−4 mr \approx 7.5 \times 10^{-4}\,\text{m}r≈7.5×10−4m

So the correct option is:

A: 7.5×10−4 m7.5 \times 10^{-4}\,\text{m}7.5×10−4m

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