A beam of electrons is moving in a horizontal circle in a uniform magnetic field that is directed downward. The speed of the electrons is 1.0*10^7 m/s and the magnitude of themagnetic field is 1.0*10^-3 T.

As seen from above, are the electrons moving clockwise or counter clockwise?

What is the radius of the elecron's orbit?

What is the period of their motion?

I just can't understand the right hand rule. if anyone wants to try to explain it to me, I'd be grateful.

To determine the direction of the electron's motion, we can apply the right-hand rule. Imagine holding your right hand perpendicular to the plane of the electrons' circular path, with your fingers pointing in the direction of the magnetic field (downward in this case). If you curl your fingers towards the direction the electrons are moving (towards the center of the circle), your thumb should point in the direction of the electrons' motion.

Using the right-hand rule in this scenario, you will find that the electrons are moving in a clockwise direction as seen from above.

To find the radius of the electron's orbit, we can first consider the equation for the centripetal force acting on a charged particle in a magnetic field:

F = qvB

Where:
F = Centripetal force
q = Charge of the electron
v = Velocity of the electron
B = Magnetic field strength

The centripetal force is given by the equation:

F = mv^2 / r

Where:
m = Mass of the electron
v = Velocity of the electron
r = Radius of the orbit

Setting the two equations equal to each other:

mv^2 / r = qvB

Simplifying, we get:

r = mv / qB

Plugging in the values:

m = mass of an electron (9.1 * 10^-31 kg)
v = velocity of the electron (1.0 * 10^7 m/s)
q = charge of an electron (-1.6 * 10^-19 C)
B = magnetic field strength (1.0 * 10^-3 T)

Calculating the radius, we find:

r = (9.1 * 10^-31 kg * 1.0 * 10^7 m/s) / (1.6 * 10^-19 C * 1.0 * 10^-3 T)

r ≈ 5.688 * 10^-4 m (or approximately 0.5688 mm)

The period of their motion can be calculated using the equation:

T = 2πr / v

Plugging in the values:

r = radius of the electron's orbit (5.688 * 10^-4 m)
v = velocity of the electron (1.0 * 10^7 m/s)

Calculating the period, we find:

T = (2 * 3.1416 * 5.688 * 10^-4 m) / (1.0 * 10^7 m/s)
T ≈ 3.583 * 10^-9 s (or approximately 3.583 nanoseconds)

Now, let's discuss the right-hand rule. The right-hand rule is a way to determine the direction of force on a charged particle moving through a magnetic field. It uses the orientation of your right hand and the three fingers to represent the direction of the magnetic field (thumb), the direction of the current or velocity (index finger), and the direction of the force (middle finger). By aligning your hand accordingly, you can determine the relative directions of these three factors.

To determine whether the electrons are moving clockwise or counterclockwise, you can use the right-hand rule. Here's how you can apply it in this scenario:

1. Extend your right hand: Keep your thumb, index finger, and middle finger perpendicular to each other, forming a right angle.

2. Point your thumb in the direction of the electron's velocity vector, which is horizontal in this case (from the information provided).

3. Point your index finger in the direction of the magnetic field vector, which is downward.

4. Your middle finger will naturally point in the opposite direction to the electron's motion.

If your middle finger is pointing counterclockwise when looking down from above, then the electron is moving counterclockwise. If it is pointing clockwise, then the electron is moving clockwise.

Now let's move on to calculating the radius of the electron's orbit and the period of their motion:

The force experienced by a moving charged particle in a magnetic field is given by the equation F = qvB, where F is the force, q is the charge of the particle, v is its velocity, and B is the magnetic field.

In this case, as the electrons move in a circle, the magnetic force provides the necessary centripetal force to keep the electrons in their circular path.

The magnetic force can be equated to the centripetal force:

qvB = m(v^2/r),

where m is the mass of the electron, v is its velocity, and r is the radius of the orbit.

Since we are looking for the radius, we can rearrange the equation to solve for r:

r = mv / (qB).

Using the provided values:
m (mass of electron) = 9.1 x 10^-31 kg,
v (velocity of electrons) = 1.0 x 10^7 m/s,
q (charge of an electron) = -1.6 x 10^-19 C,
B (magnetic field) = 1.0 x 10^-3 T,

Substituting the values in, we find:

r = (9.1 x 10^-31 kg) * (1.0 x 10^7 m/s) / (1.6 x 10^-19 C * 1.0 x 10^-3 T).

By performing the calculations, the radius of the electron's orbit is approximately equal to 5.69 x 10^-4 meters (or 0.569 mm).

To determine the period (T) of their motion, we can use the fact that the velocity is related to the circumference of the circle and the period by the equation:

v = (2πr) / T.

Rearranging for T, we get:

T = (2πr) / v.

Using the previously calculated radius (r) and the velocity (v) of the electrons, we can compute the period:

T = (2π * 5.69 x 10^-4) / (1.0 x 10^7).

By evaluating the expression, the period of the electron's motion is approximately equal to 3.58 x 10^-8 seconds (or 35.8 nanoseconds).

If you have any further questions or need more clarification, feel free to ask!