Ask questions and get helpful answers.

A particle with a charge of 0.08 C is moving at right angles to a uniform magnetic field with a strength of 0.8 T. The velocity of the charge is 800 m/s.

What is the magnitude of the magnetic force exerted on the particle?

Answer is in N

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
1 answer
  1. F = Q ( v cross B)
    cross is the cross product in general v B sin theta
    where theta is the angle between v and B
    since here it is 90 degrees, sin theta = 1
    and it is just
    F =Q v B = .08 * 800 * 0.8 Newtons

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
    👤
    Damon

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.