Question

A charged particle is moving with velocity v in a magnetic field of induction B. The force on the particle will be maximum when
A. v and B are in the same direction
B. v and B are in opposite directions
C. v and B are perpendicular
D. v and B are at an angle of \[45^\circ \]

Answer 1

Hint: If a region is occupied with magnetic field lines and a charge particle enters into the region with non-zero velocity then it experiences the magnetic force. The magnetic force on the moving charged particle is given by Lorentz force law.

Formula used:
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
here \[\vec F\] is the magnetic force vector, \[\vec v\] is the velocity of the charged particle and \[\vec B\] is the magnetic field in the region.

Complete step by step solution:
It is given that a charged particle is moving in the region with velocity v. The magnetic field induction in the region is given as B. We need to determine the condition for which the force acting on the particle is maximum. Using the Lorentz law of force, the magnetic force on the moving charged particle can be given as,
\[\vec F = q\left( {\vec v \times \vec B} \right)\]

If the angle between the velocity vector and the magnetic field vector is \[\theta \] then the magnitude of the magnetic force on the charged particle is,
\[F = qvB\sin \theta \]
For the force to be maximum, the value of sine of the angle must be maximum. As we know that the maximum value of the sine function is 1. So, for the maximum value of the magnetic force the value of sine of angle must be equal to 1, i.e. \[\theta = 90^\circ \]. Hence, the magnetic force on the particle will be maximum when the velocity and magnetic field induction is perpendicular to each other.

Therefore, the correct option is C.

Note: For the magnetic force to be minimum, the minimum magnitude of the sine of the angle is zero, i.e. the velocity and magnetic field should be parallel to each other.