Question

An electron is moving at a speed of 4.8 × 10 6 m/s at an angle of 30.0° with respect to a uniform magnetic field of 9.2 × 10 –4 T. What is the radius of the resulting helical path? ( me = 9.11 × 10 –31 kg, qe = 1.60 × 10 –19 C)

Answer 1

Answer:

r = 5.94 10⁻² m

Explanation:

The magnetic force is given by the ratio

        F =q v x B

The bold indicate vectors and the vector product determines that the force is perpendicular to the other two vectors, so the modulus of the velocity does not change, but its direction, therefore the acceleration in Newton's second law is centripetal

         F = m a

   

The magnitude of the force is

      F = q v B sin θ

The centripetal acceleration is

      a = v² / r

Let's replace

       q v B sin θ = m v² / r

       r = m v / (q B sin θ)

Let's calculate

      r = 9.11 10⁻³¹  4.8 10⁶ / (1.60 10⁻¹⁹  9.2 10⁻⁴ sin 30)

     r = 5.94 10⁻² m