Given:
The magnitude of the charge, q=3.0×10⁻⁹ C
The velocity of the charge, v=12 m/s
The magnetic field, B=8×10⁻³ T
To find:
The magnetic force.
Explanation:
The magnetic force is given by the equation,
[tex]F=qvB[/tex]
On substituting the known values,
[tex]\begin{gathered} F=3.0\times10^{-9}\times12\times8\times10^{-3} \\ =2.88\times10^{-10}\text{ N} \end{gathered}[/tex]
Final answer:
The magnetic force is 2.88×10⁻¹⁰ N
