Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 t where v and b are perpendicular. give your answer in centimetres.

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r =

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =   = 0.0328 m, or 3.28  cm.

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

                                              ⇒      

here   k is the proportionality constant whose value is 1.

                                                            [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.