A proton travels at a speed of 2.0..*.10Ò meters/second. Its velocity is at right angles with a magnetic field of strength 5.5..*.10ȳ tesla. What is the magnitude of the magnetic force on the proton?
A proton travels at a speed of 2.0..*.10Ò meters/second. Its velocity is at right angles with a magnetic field of strength 5.5..*.10ȳ tesla. What is the magnitude of the magnetic force on the proton?
To find the magnitude of the magnetic force on the proton, you can use the formula for the magnetic force on a charged particle moving through a magnetic field.
The formula for the magnetic force (F) on a charged particle is:
F = q * v * B * sin(theta)
where:
- F is the magnetic force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic field strength
- theta is the angle between the velocity vector and the magnetic field vector
In this case, the proton has a charge of +1.6 x 10^-19 coulombs (q), a velocity of 2.0 x 10^7 meters/second (v), and the magnetic field strength is 5.5 x 10^3 teslas (B). The velocity is at right angles (90 degrees) to the magnetic field, so the sine of the angle (sin(theta)) is 1.
Plugging in the given values into the formula:
F = (1.6 x 10^-19 C) * (2.0 x 10^7 m/s) * (5.5 x 10^3 T) * 1
Now we can calculate the magnetic force:
F = 1.6 x 10^-19 C * 2.0 x 10^7 m/s * 5.5 x 10^3 T
Using scientific notation, multiply the numbers:
F = 1.6 * 2.0 * 5.5 x 10^-19 * 10^7 * 10^3 N
F = 17.6 x 10^-9 N
Simplifying:
F = 1.76 x 10^-8 N
Therefore, the magnitude of the magnetic force on the proton is 1.76 x 10^-8 Newtons.