Answer:
The magnitude of the force F on a positive charge q in the case that the velocity v and the magnetic field B are perpendicular is [tex]F = q\cdot v\cdot B[/tex], measured in newtons.
Explanation:
From classical theory on Magnetism, the vectorial form of the magnetic force on a particle is given by:
[tex]\vec F = q\cdot \vec v \times \vec B[/tex] (Eq. 1)
Where:
[tex]\vec F[/tex] - Magnetic force, measured in newtons.
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in tesla.
By definition of cross product, we get that magnitude of magnetic force on a positive charge [tex]q[/tex] is:
[tex]F = q\cdot v\cdot B \cdot \sin \theta[/tex] (Eq. 2)
Where:
[tex]v[/tex] - Speed of the particle, measured in meters per second.
[tex]B[/tex] - Magnitude of the magnetic field, measured in tesla.
[tex]\theta[/tex] - Angle between the velocity of the particle and magnetic field, measured in sexagesimal degrees.
If velocity and magnetic field are perpendicular, then (Eq. 2) is reduced into this form: ([tex]\theta = 90^{\circ}[/tex])
[tex]F = q\cdot v\cdot B[/tex]