Answer:
145 degree
Explanation:
We are given that
Magnetic field=B=2 T
Charge on particle,q=[tex]-1.75\times 10^{-8} C[/tex]
Velocity, v=[tex]44.5 km/s[/tex]
Angle,[tex]\theta=55^{\circ}[/tex]
We have to find the direction of the force exerted on this particle by magnetic field with magnitude 2 T in the z direction.
Component of velocity along x- axis
[tex]v_x=vcos\theta=44.5 cos55=25.53 km/s[/tex]
Along y- axis
[tex]v_y=-44.5sin55=-36.45km/s[/tex]
[tex]\theta=tan^{-1}(\frac{-36.45}{25.53})[/tex]
[tex]\theta=tan^{-1}(-1.4277)=-55^{\circ}=180-55=145^{\circ}[/tex]
Hence, the direction of force =145 degree
