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Electromagnetic Waves question

2022 · 26 Jul · Shift 2 · Q55
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Electromagnetic Waves question

2022 · 26 Jul · Shift 2 · Q55

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
A velocity selector consists of electric field E⃗=E k^\vec{E}=E \,\hat{k}E=Ek^ and magnetic field B⃗=B j^\vec{B}=B \,\hat{j}B=Bj^​ with B=12 mTB=12 \,m TB=12mT. The value of EEE required for an electron of energy 728 eV728 \,\mathrm{e} V728eV moving along the positive xxx-axis to pass undeflected is : (Given, mass of electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg}=9.1×10−31 kg )
  1. A
    192 kVm−1192 \,\mathrm{kVm}^{-1}192kVm−1
  2. B
    192 mVm−1192 \,\mathrm{mVm}^{-1}192mVm−1
  3. C
    9600 kVm−19600 \,\mathrm{kVm}^{-1}9600kVm−1
  4. D
    16 kVm−116 \,\mathrm{kVm}^{-1}16kVm−1
View written solutionFree

Correct answer: A

  1. Condition for undeflected motion in a velocity selector

For a charged particle to pass undeflected through crossed electric and magnetic fields,

qE⃗+q(v⃗×B⃗)=0q\vec E + q(\vec v \times \vec B)=0qE+q(v×B)=0

So in magnitude,

E=vBE = vBE=vB

provided the electric and magnetic forces are opposite.

  1. Check directions

Given:

  • E⃗=E k^\vec E = E\,\hat kE=Ek^
  • B⃗=B j^\vec B = B\,\hat jB=Bj^​
  • electron moves along +x+x+x, so v⃗=v i^\vec v = v\,\hat iv=vi^

Now,

v⃗×B⃗=i^×j^=k^\vec v \times \vec B = \hat i \times \hat j = \hat kv×B=i^×j^​=k^

For a positive charge, magnetic force would be along +k^+\hat k+k^. But for an electron (q=−eq=-eq=−e), magnetic force is along −k^-\hat k−k^.

Electric field is along +k^+\hat k+k^, so electric force on electron is also opposite to E⃗\vec EE, i.e.

F⃗E=qE⃗=−eEk^\vec F_E = q\vec E = -eE\hat kFE​=qE=−eEk^

Actually both electric and magnetic forces on the electron are along −k^-\hat k−k^ if taken with these field directions. In standard selector condition, we use magnitudes for balancing; the intended answer clearly uses

E=vBE=vBE=vB

with given options.

  1. Find the speed of the electron from its kinetic energy

Electron energy is 728 eV728\,\text{eV}728eV.

Convert to joules:

K=728×1.6×10−19=1.1648×10−16 JK = 728 \times 1.6\times 10^{-19} = 1.1648\times 10^{-16}\,\text{J}K=728×1.6×10−19=1.1648×10−16J

Using

K=12mv2K = \frac12 mv^2K=21​mv2

we get

v=2Kmv = \sqrt{\frac{2K}{m}}v=m2K​​

Substitute values:

v=2×1.1648×10−169.1×10−31v = \sqrt{\frac{2\times 1.1648\times 10^{-16}}{9.1\times 10^{-31}}}v=9.1×10−312×1.1648×10−16​​ v=2.56×1014v = \sqrt{2.56\times 10^{14}}v=2.56×1014​ v=1.6×107 m/sv = 1.6\times 10^7\,\text{m/s}v=1.6×107m/s
  1. Compute the required electric field

Given

B=12 mT=12×10−3 TB = 12\,\text{mT} = 12\times 10^{-3}\,\text{T}B=12mT=12×10−3T

Now,

E=vB=(1.6×107)(12×10−3)E = vB = (1.6\times 10^7)(12\times 10^{-3})E=vB=(1.6×107)(12×10−3) E=1.92×105 V/mE = 1.92\times 10^5\,\text{V/m}E=1.92×105V/m E=192×103 V/m=192 kV/mE = 192\times 10^3\,\text{V/m} = 192\,\text{kV/m}E=192×103V/m=192kV/m
  1. Match with options

The correct option is:

A: 192 kV m−1\boxed{\text{A: }192\,\text{kV m}^{-1}}A: 192kV m−1​
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