When a charged particle moves through a perpendicular magnetic field, it undergoes a change in its

CONCEPT:

The motion of the charged particle in a uniform magnetic field:

• When a charged particle q enters a magnetic field \(\vec B\) with velocity \(\vec v\), it experiences a force

\(⇒ \vec F = q\left( {\vec v \times \vec B} \right)\)

• The direction of this force is perpendicular to both \(\vec v\) and \(\vec B\). The magnitude of this force is

⇒ F = qvB sin θ

EXPLANATION:

• The direction of this force is perpendicular to both \(\vec v\) and \(\vec B\) and the magnitude of this force is given as

⇒ F = qvB sin θ

• If a charged particle enters a magnetic field with velocity v such that the direction between the velocity of the charged particle q and magnetic field B is 90°, then it experiences a maximum force.

Here θ = 90°, so

⇒ F = qvB sin 90° = qvB = a maximum force

• As the magnetic force acts on a particle perpendicular to its velocity, it does not do any work on the particle. It does not change the kinetic energy or the speed of the particle.

• The magnetic field B is perpendicular to the paper and going into it (show by small cross).
• A charge +q is projected with speed v in the plane of the paper. The velocity is perpendicular to the magnetic field. A force F = qvB acts on the particle perpendicular to both \(\vec v\) and \(\vec B\).
• This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. Therefore option 4 is correct.