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

Answer:

a=0.212 m/s²

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

$a=\dfrac{qVB\ sin\theta}{m}$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2$

a=0.212 m/s²