A charge of 3 imes 10⁻⁶ C moves with a velocity of 2 imes 105 m/s perpendicular to a magnetic field of 0.5 T. Calculate the magnetic force on the charge.

The charge is $q = 3 \times 10^{-6}$ C.
The velocity is $v = 2 \times 10^5$ m/s. (Note: The unit in the question, m/s$^2$, is typically for acceleration. We will proceed assuming it's m/s for velocity, which is standard for this type of problem.)
The magnetic field strength is $B = 0.5$ T.
The charge moves perpendicular to the magnetic field, so the angle $\theta = 90^\circ$.

The magnetic force $F$ acting on a charge $q$ moving with velocity $v$ in a magnetic field $B$ at an angle $\theta$ is given by:

Substitute the given values into the formula:

Since $\sin(90^\circ) = 1$:

The magnetic force acting on the charge is $\text{0.3 N}$.