The electrons in the beam of a television tube have a kinetic energy of 2.10 10-15 J. Initially, the electrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 5.00 10-5 T.

(a) In what direction are the electrons deflected by this field component?

due south (correct)
due west
due east
due north

(b) What is the acceleration of an electron in part (a)?

I have tried everything and all of my answers are wrong. In part a the correct answer is due south. Please help with part (b).

Your answer for the magnitude of the acceleration will not have a negative sign. However, you must specify the direction.

You should be familiar with the term "cross prodct" if you are studying the forces of magnetic fields. It is a vector concept and vectors are usually tught in introductory calculus and analytic geometry. Awhat a cross product A X B means is |A|*|B| sin theta, where |A| amd |B| are the magnitudes of two vectors and theta is the angle between them. The direction of A X B is perpendicular to the plane containing A and B

Force = q V X B

where V and B are vectors and "x" denotes the vector "cross product". For an electron, q is negative. In (a), V is towards the east (unit vector i) and B is down (unit vector -k). The vector V x B is north (according to he right hand rule) and the minus sign of q makes the force to the south.

(b) The acceleration is F/m, in the same direction as the force F. m is the mass of the electron

I am so confused! Can you please define "cross product". I interpret it to mean that you multiply the value of V and B. Obviously this is incorrect.

Will my answer have a negative value because of the mass of the electron?

This is what I did:

Acceleration = [(q V X B)/ m
(-1.60x10^-19)(6.789932x10^12)(X)(5.00x10^-5)] / 9.11x10^-31)

I got the value of V from the equation for KE

KE = 1/2m V^2

2.10x10^-5 J = 1/2(9.11x10^-31)V^2

(2.10x10^-5)/(4.555x10^-31) = V^2

4.610318x10^25 = V^2

6.789932x10^12 = V

I suggest you learn more about vector cross products here, if a description is not in your textbook somewhere:

http://en.wikipedia.org/wiki/Cross_product

The electrons in the beam of a television tube have a kinetic energy of

2.40*10-15 J. Initially, the electrons move horizontally from west to east. The
vertical component of the earth’s magnetic field points down, toward the
surface of the earth, and has a magnitude of 2.00*10-5 T. (a) in what
direction are the electrons deflected by this field component? (b) What is the
acceleration of an electron in part (a)?

To determine the acceleration of an electron in a magnetic field, you can use the equation for the magnetic force on a charged particle:

F = q * v * B * sin(theta)

Where:
- F is the force experienced by the electron
- q is the charge of the electron (1.6 x 10^-19 C)
- v is the velocity of the electron
- B is the magnitude of the magnetic field (5.00 x 10^-5 T)
- theta is the angle between the velocity vector and the magnetic field vector

In this case, the electrons are moving horizontally from west to east, and the vertical component of the Earth's magnetic field points down toward the surface of the Earth. Therefore, the angle between the velocity vector and the magnetic field vector (theta) is 90 degrees.

Now, we need to find the velocity of the electrons (v). The kinetic energy of an object is given by:

KE = (1/2) * m * v^2

Where:
- KE is the kinetic energy (2.10 x 10^-15 J)
- m is the mass of the electron (9.11 x 10^-31 kg)

We can rearrange this equation to solve for v:

v^2 = (2 * KE) / m
v = sqrt((2 * KE) / m)

Now we can substitute the given values into the equation to find the velocity of the electrons:

v = sqrt((2 * 2.10 x 10^-15 J) / 9.11 x 10^-31 kg)

Using a calculator, we find that v ≈ 3.16 x 10^5 m/s.

Now we can calculate the acceleration:

F = q * v * B * sin(theta)
a = F / m

Since the force and the acceleration are parallel, we can simply set a = F / m:

a = (q * v * B * sin(theta)) / m

Plugging in the values:

a = (1.6 x 10^-19 C * 3.16 x 10^5 m/s * 5.00 x 10^-5 T * sin(90°)) / 9.11 x 10^-31 kg

Calculating this expression, we find that the acceleration of the electron is approximately 8.72 x 10^14 m/s².