A circular loop with a radius of 20 cm carries a 10 A current. An electron passes through the centre of the loop, crossing its plane at a speed of 2.0 10^6 m s ⁻¹ . If the electron's direction of motion is at a 30^ angle to the axis of the circle, determine the magnitude of the magnetic force acting on the electron at the moment it crosses the plane.

Step-by-step solution

The magnetic field at the centre of a circular loop is given by B = ₀ I 2R Substituting the given values: B = 4 10⁻⁷ 10 2 0.2 = 10⁻⁵ T The direction of this magnetic field is along the axis of the circular loop. The magnitude of the magnetic force acting on the electron is given by F = qvB Substituting the values of charge, velocity, magnetic field, and angle: F = (1.6 10⁻¹⁹) (2.0 10^6) ( 10⁻⁵) 30^ F = 1.6 10⁻¹⁹ 2.0 10^6 10⁻⁵ 1 2 F = 1.6 10⁻¹⁸ N = 16 10⁻¹⁹ N