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
[tex]a=\dfrac{qVB\ sin\theta}{m}[/tex]
[tex]a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2[/tex]
[tex]a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2[/tex]
a=0.212 m/s²
