Physics Circle home Student Login

Home›Chapters›Electrostatics PYQs›Motion of Charges in an Electric Field›Question 17

JEE Main 2023 (24 Jan, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2023 (24 Jan, Shift 1)

Q17JEE Main 2023 · 24 JAN · SHIFT 1Medium

Motion of Charges in an Electric Field

A stream of positively charged particles having qm=2×1011 Ckg\frac{q}{m} = 2 \times 10^{11}\,\frac{C}{kg} and velocityv⃗0=3×107i^ m/s\vec{v}_0 = 3 \times 10^{7}\hat{i}\text{ m/s} is deflected by an electric field 1.8j^1.8\hat{j} kV/m. The electric field exists in a region of 10 cm along xx direction. Due to the electric field, the deflection of the charged particles in the yy direction is _________ mm.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. ay=qmE=2×1011×1.8×103=3.6×1014 m/s2a_y = \frac{q}{m}E = 2 \times 10^{11} \times 1.8 \times 10^{3} = 3.6 \times 10^{14}\text{ m/s}^2
  2. Time in field: t=0.13×107=13×10−8 st = \frac{0.1}{3 \times 10^{7}} = \frac{1}{3} \times 10^{-8}\text{ s}
  3. y=12ayt2=12×3.6×1014×19×10−16=2×10−3 my = \frac{1}{2}a_y t^2 = \frac{1}{2} \times 3.6 \times 10^{14} \times \frac{1}{9} \times 10^{-16} = 2 \times 10^{-3}\text{ m}
  4. So the deflection is 2 mm.
Report a mistake

More questions on Motion of Charges in an Electric Field