To calculate the magnitude of the magnetic force acting on a charged particle moving in a magnetic field, we use the following equation:
\[ F = qvB\sin(\theta) \]
Where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle (in Coulombs),
- \( v \) is the velocity of the particle (in meters per second),
- \( B \) is the magnetic field strength (in Tesla),
- \( \theta \) is the angle between the direction of the velocity and the magnetic field.
In this case, you've stated that the charged particle is moving perpendicular to the magnetic field. When the motion is perpendicular, \( \theta = 90 \) degrees, and the sine of 90 degrees is 1. This simplifies the formula to:
\[ F = qvB \]
Now we can plug in the given values:
- \( q = 30 \) C (charge of the particle),
- \( v = 2 \) m/s (velocity of the particle),
- \( B = 0.5 \) T (magnetic field strength).
Using the simplified formula:
\[ F = 30 \text{ C} \times 2 \text{ m/s} \times 0.5 \text{ T} \]
\[ F = 60 \text{ C m/s} \times 0.5 \text{ T} \]
\[ F = 30 \text{ N} \]
Therefore, the magnitude of the magnetic force on the particle is 30 Newtons, which isn't one of the options you provided. Based on the choices given:
A) BY 50N (possibly a typo, and B might not be a correct choice)
B) Voltmeter (irrelevant, not a force value)
C) 5N (incorrect value)
D) 10N (incorrect value)
Since none of the options match the calculated result, there might be an issue with the choices given. The correct magnitude of the magnetic force, as calculated, is 30 N.
