After being accelerated through a potential difference of 12 kV , a charged particle attains a speed of 1.0 10^6 m s ⁻¹ . It is subsequently injected perpendicularly into a magnetic field of strength 0.2 T . Determine the radius of the circle described by the particle.

Step-by-step solution

The kinetic energy gained by the charged particle when accelerated through a potential difference V is given by: 1 2 mv^2 = qV m q = 2V v^2 When the particle is injected perpendicularly into a magnetic field B , the radius r of the circular path is: r = mv qB Substituting the value of m q into the expression for r : r = ( 2V v^2 ) v B = 2V vB Given V = 12 10^3 V , v = 1.0 10^6 m s ⁻¹ , and B = 0.2 T , we get: r = 2 12 10^3 1.0 10^6 0.2 r = 24 10^3 2 10^5 r = 12 10⁻² m = 12 cm