A velocity selector consists of electric field E = Ek and magnetic field B = Bj 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) (a) 192 kVm⁻¹ (b) 192 mVm⁻¹ (c) 9600 kVm⁻¹ (d) 16 kVm⁻¹

Answer: (a). Undeflected in a velocity selector when qE = qvB, so E = vB. v = √(2K/m) = √2 × 728 × 1.6 × 10⁻¹⁹9.1 × 10⁻³¹ = √2.56 × 10¹³ = 1.6 × 10⁷ m/s E = (1.6 × 10⁷)(12 × 10⁻³) = 1.92 × 10⁵ V/m So the answer is (a) 192 kV m⁻¹.

A velocity selector consists of electric field E = Ek and magnetic… | JEE Main 2022 (26 Jul, Shift 2)

Q59JEE Main 2022 · 26 JUL · SHIFT 2Medium

Moving Charge in a Magnetic Field

A velocity selector consists of electric field $\vec{E} = E\hat{k}$ and magnetic field $\vec{B} = B\hat{j}$ with $B = 12\text{ mT}$ . The value of $E$ required for an electron of energy $728\text{ eV}$ moving along the positive $x$ -axis to pass undeflected is:

(Given, mass of electron $= 9.1 \times 10^{-31}\text{ kg}$ )

Show answer and solution

Answer: (a)

Undeflected in a velocity selector when $qE = qvB$ , so $E = vB$ .
$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{2.56 \times 10^{13}} = 1.6 \times 10^{7}\text{ m/s}$
$E = (1.6 \times 10^{7})(12 \times 10^{-3}) = 1.92 \times 10^{5}\text{ V/m}$
So the answer is (a) $192\text{ kV m}^{-1}$ .

Report a mistake

Moving Charge in a Magnetic Field question from JEE Main 2022 (26 Jul, Shift 2)

More questions on Moving Charge in a Magnetic Field