Question

A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed of the particle is 500 m/s and its velocity vector makes an angle of 45 with the magnetic field vector. What is the magnitude of the acceleration of the particle?

Answer 1

a=0.212 m/s²

Step-by-step explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

`a=(qVB\ sin\theta)/(m)`

`a=(10^(-9)* 500* 0.003*\ sin45^(\circ))/(5* 10^(-9))\ m/s^2`

`a=(10^(-9)* 500* 0.003*(1)/(\sqrt2))/(5* 10^(-9))\ m/s^2`

a=0.212 m/s²