A ^9 Be nucleus containing four protons and five neutrons has a mass of 1.50×10−26 kg and is traveling vertically upward at 1.70 km/s.

If this particle suddenly enters a horizontal magnetic field of 1.25 T pointing from west to east, find the direction of its acceleration vector the instant after it enters the field.
north, east, west, or south?

If this particle suddenly enters a horizontal magnetic field of 1.25 T pointing from west to east, find the magnitude of its acceleration vector the instant after it enters the field.

To find the direction and magnitude of the acceleration vector, we can use the right-hand rule.

First, let's determine the direction of the acceleration vector. The direction of the acceleration experienced by a charged particle moving through a magnetic field is given by the cross product of the velocity vector and the magnetic field vector.

In this case, the charged particle is traveling vertically upward, and the magnetic field points from west to east. To find the direction of the acceleration, we can use the right-hand rule as follows:

1. Point your right thumb in the direction of the velocity vector (upward).
2. Point your fingers in the direction of the magnetic field vector (west to east).
3. Your palm will then point in the direction of the acceleration vector.

Based on the right-hand rule, the direction of the acceleration vector will be south.

Next, let's find the magnitude of the acceleration vector. The magnitude of the acceleration experienced by a charged particle moving through a magnetic field is given by the equation:

a = q * v * B / m

where:
a is the magnitude of the acceleration vector
q is the charge of the particle
v is the magnitude of the velocity vector
B is the magnitude of the magnetic field vector
m is the mass of the particle

In this case, the charge of a proton is positive and equal to +e, where e is the elementary charge. The mass of the particle is given as 1.50x10^(-26) kg, and the velocity is given as 1.70 km/s.

Using these values, we can calculate the magnitude of the acceleration vector as follows:

a = (+e) * (1.70 km/s) * (1.25 T) / (1.50x10^(-26) kg)

Note: Before performing the calculations, we need to convert 1.70 km/s to m/s.

1 km = 1000 m
1 s = 1000 ms

Therefore, 1.70 km/s = 1.70 x 1000 m/s = 1700 m/s.

Now, we can substitute the values and calculate the magnitude of the acceleration:

a = (+e) * (1700 m/s) * (1.25 T) / (1.50x10^(-26) kg)

The magnitude of the acceleration will depend on the value of the elementary charge, which is approximately 1.602x10^(-19) C.

Therefore, the final step is to perform the calculation:

a = (+1.602x10^(-19) C) * (1700 m/s) * (1.25 T) / (1.50x10^(-26) kg)

By performing the calculation, you will find the magnitude of the acceleration vector.