Verified Solution • 2025–26 Session moderate

QJEEM22Ph5163: JEE Main 2022 Physics - Class 12 JEEM22Ph5163 Solution: A velocity selector consists of electric field and magnetic…

A velocity selector consists of electric field E⃗=E k^\vec{E}=E \,\hat{k} and magnetic field B⃗=B j^\vec{B}=B \,\hat{j} with B=12 mTB=12 \,m T . The value of EE required for an electron of energy 728 eV728 \,\mathrm{e} V moving along the positive xx -axis to pass undeflected is :

(Given, mass of electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg} )

Multiple Choice Options

  • A 192 kVm−1192 \,\mathrm{kVm}^{-1}
  • B 192 mVm−1192 \,\mathrm{mVm}^{-1}
  • C 9600 kVm−19600 \,\mathrm{kVm}^{-1}
  • D 16 kVm−116 \,\mathrm{kVm}^{-1}

Step-by-Step Verified Solution

v=EBv = {E \over B} and K=12mv2K = {1 \over 2}m{v^2}

⇒2Km×B=E\Rightarrow \sqrt {{{2K} \over m}} \times B = E

⇒E=2×728×1.6×10−199.1×10−31×12×10−3\Rightarrow E = \sqrt {{{2 \times 728 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}} \times 12 \times {10^{ - 3}}

=192000= 192000 V/m

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