Question

21. Suppose a 3.0x10°C charge is moving East with a velocity of 12m/s, through amagnetic field, with a strength of 8x10-3T. Find the value of the magnetic force on thecharge. (Assume your angle is 90 degrees.)for

Answer 1

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.

Step-by-step explanation:

The magnetic force is given by the equation,

`F=qvB`

On substituting the known values,

`\begin{gathered} F=3.0*10^(-9)*12*8*10^(-3) \\ =2.88*10^(-10)\text{ N} \end{gathered}`

Final answer:

The magnetic force is 2.88×10⁻¹⁰ N