In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge +5e that are traveling perpendicular to the fields at 8.75 km/s. The magnetic field availabe to you has a magnitude of 0.550 T. (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions. (c) Will your velocity selector also allow the following ions (having the same velocity as the +5e ions) to pass through undeflected? (i) negative ions of charge -5e? (ii) positive ions of charge different from +5e?

ok totally lost! please help!

I assume the velocity selector leaves a certain velocity ion undeflected by having the eE force equal and opposite to the qVB force. Since the E and B fields are perpendicular, the resulting forces can be in the same direction.

(a) The E field you want is given by
E = V B

(b) E and B should both be perpendicular to the V direction as well as each other. In that way, the electric and magnetic forces act to change the beam direction only, and maximum deflection of unwanted velcoities is achieved..

(c) The size and sign of q does not matter. E and B forces will balance for all q values if the velocity is the value one wishes to select.

Sure, I can help you with that!

(a) To start, we need to understand the concepts behind the velocity selector. In a velocity selector, we can use perpendicular electric and magnetic fields to select only charged particles with a specific velocity. For a positive ion with charge +5e, to calculate the magnitude of the required electric field (E), we can use the following formula:

E = (B * v) / (q)

Where:
- E is the magnitude of the electric field
- B is the magnitude of the magnetic field
- v is the velocity of the charged particle
- q is the charge of the particle

In this case, B = 0.550 T, v = 8.75 km/s = 8750 m/s, and q = +5e, where e is the elementary charge (1.6 x 10^-19 C).

Plugging these values into the formula, we get:

E = (0.550 T * 8750 m/s) / (5 * 1.6 x 10^-19 C)

Calculating this, we find that the magnitude of the required electric field is approximately 1.078 x 10^7 V/m.

(b) To determine the orientation of the fields relative to each other and to the velocity of the ions, we can use the right-hand rule. If we imagine the velocity of the ions to be directed into the page (perpendicular to the electric and magnetic fields), the direction of the magnetic field can be represented by using your right hand. When you point your thumb in the direction of the velocity, your fingers will curl in the direction of the magnetic field. The electric field should be perpendicular to both the magnetic field and the velocity, so its orientation will depend on the specific design of your velocityselector device.

(c) Now let's consider the additional ions:

(i) For the negative ions of charge -5e, the same formula can be used to find the magnitude of the electric field they experience. Plugging in the values of B, v, and q, we will get the electric field E_neg. Since the charge is negative, the electric field will have the opposite direction compared to the positive ions. So, the negative ions will experience the opposite force and will not pass through undeflected.

(ii) For positive ions of charge different from +5e, the electric field required to keep them undeflected will be different. To calculate the required electric field in this case, we need to know the specific charge value (q) of the ions. Using the same formula as before, with the new charge value and the same velocity and magnetic field, we can calculate the new electric field magnitude (E_diff). If E_diff is not equal to the magnitude of the original electric field (E), these ions will not pass through undeflected.

I hope this helps clarify the steps and concepts involved in designing a velocity selector using uniform perpendicular electric and magnetic fields! Let me know if you have any further questions.

To design a velocity selector using uniform perpendicular electric and magnetic fields, we can use the equation for the force on a charged particle in a magnetic field:

F = q(v x B),

where F is the force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength.

(a) To select positive ions of charge +5e traveling at 8.75 km/s, we need to find the magnitude of the electric field E.

The force on the positive ions must cancel out the magnetic force to make them travel in a straight line without deviation. Therefore, we have:

F_electric = F_magnetic.

Using the equation for the force on a charged particle in an electric field:

F_electric = qE,

where E is the electric field strength, we can equate the two forces:

qE = q(v x B).

Dividing both sides by q, we get:

E = v x B.

Substituting the values into the equation, we have:

E = (8.75 km/s) × (0.550 T).

Converting the velocity to m/s:

E = (8.75 × 10^3 m/s) × (0.550 T).

Calculating the magnitude of the electric field, we get:

E = 4.8125 × 10^3 V/m.

Therefore, the electric field magnitude needed is 4.8125 × 10^3 V/m.

(b) The two fields should be oriented perpendicular to each other and also perpendicular to the direction of velocity of the ions. This means that the electric field and magnetic field vectors should form a 90-degree angle with each other and with the direction of the ions' velocity.

(c) The velocity selector will not allow the following ions (having the same velocity as the +5e ions) to pass through undeflected:

(i) Negative ions of charge -5e: The force on them will be in the opposite direction because of their negative charge. Hence, they will be deflected in the opposite direction and will not pass through undeflected.

(ii) Positive ions of charge different from +5e: The force on these ions will be different because of the difference in charge. Therefore, they will experience a different force and will be deflected, so they will not pass through undeflected.