A particle of charge 2.5x10(^-6)C is moving at 1000 m/sec in a direction 30 degrees North of east. At the location of the particle the magnetic field has size 0.2 T. The particle is experiencing an instantaneous magnetic force (due to this sheild) of 1.71x10(^-4) newtons directed vertically down. The size of the angle A (in deg) between the velocity and the magnetic field at the location of the particle is most nearly:
0
12.6
15.9
17.1
20.0
24.1
28.3
70
To find the angle A between the velocity and the magnetic field at the location of the particle, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:
F = qvBsinθ
Where:
F = Magnetic force on the particle (1.71x10^(-4) N)
q = Charge of the particle (2.5x10^(-6) C)
v = Velocity of the particle (1000 m/s)
B = Magnetic field strength (0.2 T)
θ = Angle between the velocity and the magnetic field
Rearranging the formula to solve for θ:
θ = arcsin(F / (qvB))
Let's substitute the values into the formula and calculate θ:
θ = arcsin((1.71x10^(-4) N) / ((2.5x10^(-6) C) * (1000 m/s) * (0.2 T)))
Using a calculator, the value of θ comes out to be approximately 15.9 degrees.
Therefore, the size of the angle A (in degrees) between the velocity and the magnetic field at the location of the particle is most nearly 15.9 degrees. The closest option is 15.9.