# A particle with a mass of 6.64 * 10^-27 kg and a charge of +3.2 * 10^-19 C is accelerated... (multiple choice)

A particle with a mass of 6.64 * 10^-27 kg and a charge of +3.2 * 10^-19 C is accelerated from rest through a potential difference of 2.45 * 10^6 V. The particle then enters a uniform 6.3 T magnetic field. If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle?

A) zero Newtons

B) 1.14 * 10^-10 N

C) 6.55 * 10^-10 N

D) 3.1 * 10^-11 N

E) 7.87 * 10^-12 N

### 4 Answers

- JicotilloLv 61 decade agoFavorite Answer
Zero force? No way!

The force exerted by a magnetic field on a moving, charged particle is given by.

F = Bqv sin θ,

where F is in N, B in T, q in C, v in m/s. All data is known except "v"; since the problem states that velocity is perpendicular to the magnetic field, then θ = 90°, or θ = -90°. Either way, sin θ =1. Velocity can be ascertained from the remaining data given.

Thus,

Work done on particle = Kinetic energy gained

or

½mv² = Vq,

so

v = √(2Vq/m).

From the values given, v = 15.37 E6 m/s, and the product Bqv evaluates as 6.3 × 3.2 E-19 × 15.37 E6 = 3.099 E-11 N. The right answer is therefore D). The particle is trapped into a circular path, 5 cm (2 in) radius, and remains there until the magnetic field is cut off.

- boshellLv 43 years ago
we are able to be needing f = (m(v^2))/r to locate the radius also F = Bev (v = speed, B= Magnetic field ability, e = cost ,F = pressure) So, F = 2.40-one x 3.29^-22 x 2.4 x 10^ 8 = a million.9 x 10^-13N (2.28 x 10^-22 x( 2.4 x 10^8)^2)/F = r =6912000M = 69120Km= answer d ;)