AP Physics 2: 5.1 Magnetic Systems – Exam Style questions with Answer- FRQ

Question

In a mass spectrometer, singly charged 16O ions are first accelerated electrostatically through a voltage V to a speed vo. They then enter a region of uniform magnetic field B directed out of the plane of the paper.
a. The 16O ions are replaced with singly charged 32S ions of twice the mass and the same charge. What will be their speed in terms of vo for the same accelerating voltage?
b. When 32S is substituted for 16O in part (a), determine by what factor the radius of curvature of the ions’ path in the magnetic field changes.

▶️Answer/Explanation

Ans:

a) Since the ions have the same charge, the same work (Vq) will be done on them to accelerate them and they will gain the same amount of K as they are accelerating. Set the energies of the two ions equal.
Ko = Ks                  ½ mo vo2 = ½ (2mo)vs2            vs = vo / √2
b) In the region of the magnetic field, apply Fnet(C) = mv2/r …                    qvB = mv2/ r …                     r = mv / Bq
For the O ion                           For the S ion
r1 = movo / Bq                          r2 = (2mo)(vo / √2) / Bq                           comparing the two. R2 = ( 2 /√2 ) R1

Question

An ion of mass m and charge of known magnitude q is observed to move in a straight line through a region of space in which a uniform magnetic field B points out of the paper and a uniform electric field E points toward the top edge of the paper, as shown in region I above. The particle travels into region II in which the same magnetic field is present, but the electric field is zero. In region II the ion moves in a circular path as shown.
(a) Indicate on the diagram below the direction of the force on the ion at point P2 in region II.

(b) Is the ion positively or negatively charged? Explain clearly the reasoning on which you base your conclusion.
(c) Indicate and label clearly on the diagram below the forces which act on the ion at point P1 in region I.

(d) Find an expression for the ion’s speed v at point P1 in terms of E and B.

▶️Answer/Explanation

Ans:

a) Arrow should point radially inwards
b) Since the LHR gives the proper direction for F, v, B the charge is negative
c) Force Fe should points down (E field pushes opposite on – charges) and Fb should point up.
d) To move horizontally, Fnet = 0 … Fe = Fb …                     Eq = qvB …                       v = E/

 Question

An electron is accelerated from rest through a potential difference of magnitude V between infinite parallel plates P1 and P2. The electron then passes into a region of uniform magnetic field strength B which exists everywhere to the right of plate P2. The magnetic field is directed into the page.

a. On the diagram above, clearly indicate the direction of the electric field between the plates.
b. In terms of V and the electron’s mass and charge, determine the electron’s speed when it reaches plate P2.
c. Describe in detail the motion of the electron through the magnetic field and explain why the electron moves this way.
d. If the magnetic field remains unchanged, what could be done to cause the electron to follow a straight-line path to the right of plate P2?

▶️Answer/Explanation

Ans:

a) The E field points left since it’s a negative charge and is moved opposite the E field.
b) Work done by the accelerating plate = kinetic energy gained. W = K … Vq = ½ mv2 … v = √ (2Ve/m)
c) Using the LHR, the force on the electron would be down when it enters the B field. This will turn the charge and the resulting B force will act as a centripetal force making the charge circle.
d) Using an E field to create a force equal and opposite to the Fb could make the charge move in a straight line.
Since the charge is negative and the initial Fb is down, The E field would point down to make an upwards Fe

Scroll to Top