Magnetic Force of a Charge Moving in a Magnetic Field

A magnetic force field exerts a force on a charge in a magnetic field and these charges in motion constitute an electric current. 

If the strength of the magnetic field called flux densityDensity is the measurement of how tightly a material is packed together i.e. how closely the particles are packed in the material. The tighter the material is packed the more its... More can be represented by a vector quantityVector quantities are quantities with magnitude and direction. Examples of vector quantities include displacement, velocity, position, force, and torque. More called \( \scriptsize \overrightarrow{B}\) and the angle between the magnetic field and the direction of the charged motion is θ, when a force \( \scriptsize \overrightarrow{F}\) is applied at velocity \( \scriptsize \overrightarrow{V}\) , then,

\( \scriptsize \overrightarrow {F} = q \overrightarrow{V}\overrightarrow{B} sinθ\)

• F = force in Newton
• V = average velocity of the charge in ms^-1
• B = flux induction or magnetic induction in Tesla (T)
• q = charge in coulombs

1 Tesla = 1 Weber per m² (Wm^-2)

The expression BVsinθ can be represented as B × V

B × V = BVsin θ

 F = qVBsin θ = q (V × B)

• When V and B are parallel in the same direction, θ = 0, then F = 0
• When V and B are perpendicular, sin θ = sin 90 = 1, then 

F = qVB

Example 2.5.1:

Find the magnetic force experienced by an electron projected into a magnetic field of flux density 20 Tesla with a velocity of 4 × 10⁶ ms^-1  and in a direction of:

(i) 90°
(ii) 60°
(iii) parallel to the magnetic field

(charge on an electron = 1.6 × 10^-19 C)

Solution 

(i) F = qVB = qVB sin θ

= 1.6 x 10^-19 × 4 × 10⁶ × 20 × sin 90

= 1.6 × 10^-19 × 4 × 10⁶ × 20

= 6.4 × 10^-13 × 20

= 1.28 × 10^-11 N

(ii) F = qVB sin 60

= 1.6 × 10^-19 × 4 × 10⁶ × 20 × sin 60

= 6.4 × 10^-13 × 20 × 0.866

= 1.28 × 10^-11 × 0.866

= 1.11 × 10^-11 N

(iii) When V and B are parallel, θ = 0

∴ Sin θ = 0

∴ F = qVB sin 0

= 0