Question

An electron and a proton moving with same velocity v enter into a uniform perpendicular magnetic field. Then

• A. proton alone moves in straight line path
• B. electron alone moves in straight line path
• C. both moves in straight line path
• D. both moves in elliptical line path
• E. both moves in circular path

Answer 1

Answer: (E)

When charged particles move in a magnetic field, they experience a magnetic force that is perpendicular to both their velocity and the magnetic field. This force causes the charged particles to move in circular paths. The magnetic force on a moving charged particle is given by: \[ F = q v B \] where:
\( F \) is the magnetic force,
\( q \) is the charge of the particle,
\( v \) is the velocity of the particle, and
\( B \) is the magnetic field strength.

Since the proton and electron both have charges of equal magnitude (\( +e \) and \( -e \) respectively), and they are moving with the same velocity in the same magnetic field, they will both experience a centripetal force that causes them to move in circular paths. The radius of the circular path depends on the mass and charge of the particle, but both will follow circular trajectories. Thus, both the electron and the proton move in circular paths in the magnetic field. 

Hence, the correct option is (E) : both moves in circular path

Answer 2

When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a centripetal force and moves in a circular path. 
 
Since both the electron and the proton are charged particles and are entering with the same velocity into a uniform perpendicular magnetic field, both will experience a magnetic force given by: 

\( F = qvB \) (where \( q \) is charge, \( v \) is velocity, and \( B \) is magnetic field strength) 

This force acts as the centripetal force, making the particles move in a circular path. 

Note: The radius of the circular path will be different due to their different masses and charges, but the nature of the path (circular) remains the same. 

Final Answer: \( \boxed{\text{both moves in circular path}} \)