JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Magnetic Effects of Current

A velocity selector consists of electric field

E = E \hat{k}

and magnetic field

B = B \hat{j}

with

B = 12

mT. The value 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 \times 1 0^{- 31}

kg)

Answer: (A)

Solution

For an electron to pass undeflected through a velocity selector,

E = v B

. First, find the velocity from the kinetic energy:

\frac{1}{2} m v^{2} = 728

= 728 \times 1.6 \times 1 0^{- 19}

J. Solving:

v^{2} = \frac{2 \times 728 \times 1.6 \times 1 0 ^{- 19}}{9.1 \times 1 0 ^{- 31}}

, giving

v = 16 \times 1 0^{6}

m/s. Then

E = v B = 16 \times 1 0^{6} \times 12 \times 1 0^{- 3} = 192 \times 1 0^{3}

V/m

= 192

kVm

^{- 1}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option