A charged particle of 2 μC, accelerated by a potential difference of 100 V, enters a uniform magnetic field of magnitude 4 mT at right angle to the field. It completes a semicircle of radius 3 cm inside the magnetic field. The mass of the charged particle is _____ × 10⁻¹⁸ kg. Image
Physics · Circular Motion
JEE Mains 1 February 2023 Shift 1 Morning Shift

Question

A charged particle of 2 μC, accelerated by a potential difference of 100 V, enters a uniform magnetic field of magnitude 4 mT at right angle to the field. It completes a semicircle of radius 3 cm inside the magnetic field. The mass of the charged particle is _____ × 10⁻¹⁸ kg. Image

Physics · Circular Motion

JEE Mains 1 February 2023 Shift 1 Morning Shift

G1 Physics · Accepted Answer

Energy gain qV = (1/2)mv² and circular radius r = mv/(qB). Eliminating v gives m = qB²r²/(2V). Substituting q=2×10⁻⁶ C, B=4×10⁻³ T, r=3×10⁻² m and V=100 V gives m = 144×10⁻¹⁸ kg.