JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
A velocity selector consists of electric field E⃗=E k̂ and magnetic field B⃗=B ĵ with B=12 m T. The value of E required for an electron of energy 728 e V moving along the positive x-axis to pass undeflected is :
(Given, mass of electron =9.1 × 10⁻³¹ kg )
- (a) 192 kVm⁻¹
- (b) 192 mVm⁻¹
- (c) 9600 kVm⁻¹
- (d) 16 kVm⁻¹
Answer and solution
Answer (a) 192 kVm⁻¹
Solution
Physics sir · 4 steps
Undeflected means the two forces cancel, so E = vB. Get v from the kinetic energy first.
- Balance the forces. The electron passes straight through only when the electric force eE equals the magnetic force evB in size. The charge cancels, leaving E = vB.
- Speed from energy. The energy is 728 × 1.6 × 10⁻¹⁹ = 1.1648 × 10⁻¹⁶ J. Use 1/2mv² = K to find the speed.
v = √(2K/m) = √((2 × 1.1648 × 10⁻¹⁶)/(9.1 × 10⁻³¹)) = √(2.56 × 10¹⁴) = 1.6 × 10⁷ m s⁻¹
- Multiply by B. With B = 12 × 10⁻³ T, E = (1.6 × 10⁷)(12 × 10⁻³) = 1.92 × 10⁵ V m⁻¹.
- So. E = 192 kV m⁻¹, which is option (a).
The idea to keep. In any velocity selector, the particle that goes straight has v = E/B, whatever its charge or mass. Mass only enters when you have to turn a kinetic energy into a speed.
Answer from the NTA final answer key. Solution by PrepWiser.