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A velocity selector consists of electric field E⃗ = Ek̂ and magnetic field B⃗ = Bȷ̂ with B = 12 mT. The value of E required for an electron of energy 728 eV moving along the positive x-axis to pass undeflected is (Given: mass of electron = 9.1 × 10⁻³¹ kg)

Asked in JEE Main 26th July 2nd Shift 2022 · Crossed electric and magnetic fields

Answer: (1) 192 kV m⁻¹

Step-by-step solution

v = √(2K)/m = √(2 × 728 × 1.6 × 10⁻¹⁹)/(9.1 × 10⁻³¹) = 1.6 × 10⁷ m/s.

Undeflected: eE = evB, so E = vB.

E = 1.6 × 10⁷ × 12 × 10⁻³ = 1.92 × 10⁵ V/m = 192 kV/m.

Why the other options are wrong

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