Cambridge International AS & A Level · 20 · structured

Charged-particle forces and a velocity selector

Charged-particle forces and a velocity selector · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand: demanding
Marks: 9
Topics: 3
Answer: Complete

structured9 marks

A particle with charge +1.60×10⁻¹⁹ C moves horizontally right at 2.00×10⁶ m s⁻¹ through a magnetic field of 0.0600 T directed into the page. Electric and gravitational forces are initially absent.

(a)
Determine the magnetic force's direction and magnitude. State the direction for a negative particle with the same speed and charge magnitude.
3 marks
(b)
An electric field is now added so that the positive particle travels undeflected. State its direction, calculate its magnitude and derive the selected speed formula.
4 marks
(c)
With the fields unchanged, explain the deflection of a faster positive particle and whether a negative particle moving right at the selected speed is also undeflected.
2 marks

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Answer overview(a) Positive particle: upward, 1.92×10⁻¹⁴ N; negative particle: downward. (b) E downward with magnitude 1.20×10⁵ V m⁻¹; qE=qvB gives v=E/B. (c) Faster positive particle deflects upward; the selected-speed negative particle is also undeflected because both forces reverse.

(a)

3 marks

Determine the magnetic force's direction and magnitude. State the direction for a negative particle with the same speed and charge magnitude.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

1
Use F=Bqv sinθ with θ=90°.
2
F=0.0600×1.60×10⁻¹⁹×2.00×10⁶=1.92×10⁻¹⁴ N.
3
For positive charge v right and B into the page give force upward; reversing charge reverses force.

(b)

4 marks

An electric field is now added so that the positive particle travels undeflected. State its direction, calculate its magnitude and derive the selected speed formula.

How to approach it

1
To cancel the upward magnetic force, the electric force on the positive charge must point downward, so E is downward.
2
Balance qE=qvB.
3
Cancel q to obtain v=E/B.
4
E=vB=2.00×10⁶×0.0600=1.20×10⁵ V m⁻¹.

(c)

2 marks

With the fields unchanged, explain the deflection of a faster positive particle and whether a negative particle moving right at the selected speed is also undeflected.

How to approach it

Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.

1
For higher v the magnetic force qvB grows while electric force qE is unchanged, leaving an upward resultant.
2
At the selected speed, changing the sign of q reverses both opposing forces, so they still cancel.

Complete marking guidance

Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.

(a) Use F=Bqv sinθ with θ=90°.
(a) F=0.0600×1.60×10⁻¹⁹×2.00×10⁶=1.92×10⁻¹⁴ N.
(a) For positive charge v right and B into the page give force upward; reversing charge reverses force.
(b) To cancel the upward magnetic force, the electric force on the positive charge must point downward, so E is downward.
(b) Balance qE=qvB.
(b) Cancel q to obtain v=E/B.
(b) E=vB=2.00×10⁶×0.0600=1.20×10⁵ V m⁻¹.
(c) For higher v the magnetic force qvB grows while electric force qE is unchanged, leaving an upward resultant.
(c) At the selected speed, changing the sign of q reverses both opposing forces, so they still cancel.

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