A charged particle of mass m = 2 10⁻¹⁶ kg and charge q = 1 C is accelerated from rest through a potential difference of 100 V. It then enters a region containing a uniform magnetic field of magnitude B = 0.1 T. The angle between the particle's velocity and the magnetic field is 60^ . The pitch of the resulting helical path is x mm. The value of x is ________.

Step-by-step solution

Given: Mass, m = 2 10⁻¹⁶ kg Charge, q = 1 C = 10⁻⁶ C Potential difference, V = 100 V Magnetic field, B = 0.1 T Angle, = 60^ Using the work-energy theorem, the kinetic energy acquired by the particle is: 1 2 mv^2 = qV v = 2qV m Substituting the values: v = 2 10⁻⁶ 100 2 10⁻¹⁶ = 10¹² = 10^6 m/s The component of velocity parallel to the magnetic field is: v_ = v (60^ ) = 10^6 1 2 = 0.5 10^6 m/s The time period of one revolution is: T = 2 m qB T = 2 2 10⁻¹⁶ 10⁻⁶ 0.1 = 4 10⁻¹⁶ 10⁻⁷ = 4 10⁻⁹ s The pitch of the helical pat