A 5 mg particle carrying a charge of 5 10⁻⁶ C is moving with velocity of (3 i +2 k ) 10⁻² m/s in a region having magnetic field B = 0.1 k Wb/m ^2 . It moves a distance of meter along k when it completes 5 revolutions. The value of is ________.

Step-by-step solution

Given: Mass of the particle, m = 5 mg = 5 10⁻⁶ kg Charge, q = 5 10⁻⁶ C Velocity, v = (3 i + 2 k ) 10⁻² m/s Magnetic field, B = 0.1 k Wb/m ^2 The velocity component parallel to the magnetic field is v_ = 2 10⁻² m/s . The time period of one revolution is given by: T = 2 m qB Substituting the given values: T = 2 5 10⁻⁶ 5 10⁻⁶ 0.1 = 2 0.1 = 20 s The time taken to complete 5 revolutions is: t = 5T = 5 20 = 100 s The distance moved along the k direction (pitch for 5 revolutions) is: = v_ t = 2 10⁻² 100 = 2 m Answer: 2