Final answer:
The center of the circular path of a charged particle moving at a 45° angle to the x-axis in a uniform magnetic field directed along the z-axis is at coordinates (+mv/qB√2, -mv/qB√2, 0), making option a the correct one.
Step-by-step explanation:
The question deals with the motion of a charged particle in a uniform magnetic field when it is projected at an angle to the field. According to the information provided, the magnetic field is along the positive z-direction and the particle is launched in the x-y plane at a 45° angle to the positive x-axis. When a charged particle moves in a magnetic field, it experiences a magnetic force that is perpendicular to both the velocity of the particle and the magnetic field. This force acts as a centripetal force and causes the particle to move in a circular path.
The radius of this circular path is given by the formula r = mv/qB, where m is the mass of the particle, v is the velocity, q is the charge of the particle, and B is the magnetic field. Here, since the velocity makes a 45° angle with the x-axis, the charges x and y components of velocity are both v/√2. Therefore, the center of the circular path would be at a distance r from the origin along the -y direction (since the force will direct it to move in a circle centered below the x-axis) and along the +x direction (since it has a component of velocity in the positive x-direction).
Given the above, the coordinates of the center of the circular path would be (mv/qB√2, -mv/qB√2, 0), which makes option a (+mv/qB√2, -mv/qB√2, 0) the correct answer. The particle initially has a velocity component along the x-axis, which will remain unchanged throughout its motion because there is no force acting in the x-direction (the magnetic force acts only perpendicular to the velocity and the magnetic field).