Test: Lorentz Force(4 Dec) - JEE MCQ

4. 0.1063 m

As the electron is moving along the direction of the magnetic field, it will experience no magnetic force, but due to an electric force acting on it opposite to the direction of electric field (as it is a negatively charged particle) the velocity of the electron will decreases.

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule . The force is given by the charge times the vector product of velocity and magnetic field.

The force will have a magnitude F=qvB sin q, thus it will be maximum if sin q is maximum. Thus, angle between velocity and magnetic field should be 90^o or the charge particle moves perpendicular to the velocity vector.

Frequency (v) = qB/2πm.
Frequency is independent of radius of trajectory of particle and speed of particle.
 

The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 * 4) = 28 units.

When a charged particle enters a magnetic field B its kinetic energy remains constant as the force exerted on the particle is:

is perpendicular to so the work done by =0. This does not cause any change in kinetic energy.

Kinetic energy of a charged particle,

Radius of the circular path of a charged particle uniform magnetic field is given by

Mass of a proton, m_p = m

Mass of an α-particle. m_α = 4m

Charge of a proton, q_p = e

Charge of an α -particle, q_α = 2_e