Question

When a charged particle, moving with a velocity $v$ , is subjected to a magnetic field, the force on it is non-zero. This implies that:
A. Angles between them is either ${0^o}$ or ${180^o}$
B. Angles between them is necessarily ${90^o}$
C. Angles between them can have any value other than ${90^o}$
D. Angles between them can have any value other than ${0^o}$ and ${180^o}$

Answer 1

Hint: In this question, the charged particle is moving but the magnetic field is constant with time. Also when there is a cross product between two vectors, it can be written as the product of the magnitude of two vectors and the angle between them. If $\overrightarrow A $ and $\overrightarrow B $ are two vectors having an angle $\theta $ between them then the cross product is given by:
$\overrightarrow A \times \overrightarrow B = AB\sin \theta $

Complete step by step solution:
Electric and magnetic forces tend to change the trajectory of the charged particles but both have different ways to do it.
The force due to an electric field on a charged particle is given by:
${F_e} = q\overrightarrow E $
Where $q = $ a charge of the particle
$\overrightarrow E = $ electric field vector
This shows that the force due to the electric field is parallel to the electric field vector if the charge is positive and anti-parallel if the charge is negative. However, it doesn’t depend on the velocity of the charged particle. Therefore, the trajectory is a straight line in the case of a uniform electric field.
On the other hand, the force due to the magnetic field on a charged particle is :
$\overrightarrow {{F_m}} = q\overrightarrow v \times \overrightarrow B $
$ \Rightarrow {F_m} = qvB\sin \theta $
Where $q = $ a charge of the particle
$\overrightarrow v = $ the velocity of the charged particle
 $\overrightarrow B = $ magnetic field vector
$\theta = $ angle between the two vectors i.e. velocity vector and magnetic field vector
The magnetic force on the charged particle is perpendicular to the plane containing the magnetic field vector and velocity vector and depends on the velocity of the charged particle. Hence, it has only a deflection effect which implies that it will only change the direction of the velocity of the charged particle without changing the magnitude. Therefore, the charged particle moves in a circular motion. The right-hand rule can be used to find the direction of the magnetic field.
Now, the question says that the force should be non-zero. So according to the formula,
${F_m} = qvB\sin \theta $
The magnetic force${F_m}$ is zero only when $\theta = {0^0}$ or $\theta = {180^o}$ because $\sin {0^o} = 0$ and $\sin {180^o} = 0$ . So, to have a non-zero magnetic force, the angles between the vectors can have any value other than ${0^o}$ and ${180^o}$ .
If the magnetic force is zero, it implies that the velocity vector and the magnetic field vector are parallel to each other.
Therefore, option D is correct.

Note:
When both electric force and a magnetic force acts on a charged particle, then their combined effect is known as Lorentz force. The work done by a magnetic force is zero. To get the magnetic field lines, two poles are required. Magnetic monopoles don't exist in nature.