A proton in a cyclotron is moving with a speed of 2.97×107 m/s in a circle of radius 0.568 m. 1.67 × 10−27 kg is the mass of the proton, and 1.60218 × 10−19 C is its fundamental charge. What is the magnitude of the force exerted on the proton by the magnetic field of the cyclotron? Answer in units of N.

B = 0.546 T,  F = 2.59 10⁻¹² N

Explanation:

The magnetic force is

            F = q v x B

We can calculate the magnitude of the force and find the direction by the right hand rule

          F = q v B sin θ

Let's use Newton's second law

         F = m a

Acceleration is centripetal

         a = v² / r

We substitute

       q v B sin θ = m v² / r

The angle between the field and the radius of the circle is 90º so sin 90 = 1

        q B = m v / r

        B = m v / q r

Let's calculate ’

       B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)

        B = 0.546 T

The foce is

         F = q v B

         F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546

         F = 2.59 10⁻¹² N

the answer would be a

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures