Answer:
-(0.330m/s² ) kˆ
Explanation:
given data:
Mass of particle 'm'= 1.81 x [tex]10^{-3}[/tex] kg
Velocity 'v'= (3.00 x[tex]10^{4}[/tex] m/s)j
Charge of particle 'q'= 1.22 x [tex]10^{-8}[/tex] C
Uniform magnetic field 'B' = (1.63iˆ + 0.980jˆ )T
In order to calculate particle's acceleration, we'll use Newton's second law of motion i.e F=ma
Also,the force a magnetic field exerts on a charge q moving with velocity v is called the magnetic Lorentz force. It is given by:
F = qv × B
F= ma = qV x B
a= [tex]\frac{q(v*B)}{m}[/tex] --->eq(1)
Lets determine the value of (v x B) first
v x B= (3.00 x[tex]10^{4}[/tex] m/s)j x (1.63iˆ + 0.980jˆ )
v x B= 4.89 x [tex]10^{4}[/tex]
Plugging all the required values in eq(1)
a= [1.22 x [tex]10^{-8}[/tex] x (4.89 x [tex]10^{4}[/tex]kˆ)] / 1.81 x [tex]10^{-3}[/tex]
a=  -(0.330m/s² ) kˆ
-ve sign is representing the opposite direction
