Magnetic Fields AP Physics 2 FRQ – Exam Style Questions etc.
Magnetic Fields AP Physics 2 FRQ
Unit 12: Magnetism and Electromagnetic Induction
Weightage : 15–18%
Exam Style Practice Questions , Magnetic Fields AP Physics 2 FRQ
Question
A sample of radioactive sources is enclosed within a lead-shielded container with a narrow exit aperture that ensures that any ejected particles will exit the container directly to the right. The ejected particles pass between charged parallel conductive plates and a region of magnetic field that is directed into the page. Three particles exit the container and follow paths as shown in the figure. Students observing the particles make these statements:
Student A: Both particles 2 and 3 pass through the capacitor region undeflected so there is no net force on them. They both must be neutral particles.
Student B: The path of particle 3 implies it has a negative charge. Therefore, the bottom plate of the capacitor must be negatively charged.
Student C: Particle 1 is positive because it curves upward.
(A) List all parts of the students’ statements that are correct. Explain your reasoning for each.
(B) List all parts of the students’ statements that are incorrect. Explain your reasoning for each.
( C) On the figure, sketch the electric field vectors between the capacitor plates that are consistent with the motion of the particles.
(D) Particle 3 is detected at distance D from its exit point from the capacitor plates. Using this information, derive an expression for the charge to mass ratio (q/m) of particle 3 in terms of D, E (the electric field between the capacitor plates), and B (the magnetic field).
(E) Using the equation derived in (D), a charge to mass ratio (q/m) of 1.76 x \(10^{11}\) C/kg was calculated. Explain how this number could be used to determine the set of possible masses of particle 3.
(F) One of the radioactive sources enclosed in the shielded container is fermium-257, which has 100 protons and a half-life of 100.5 days. Fermium transmutes into Californium (Cf) by emitting an alpha particle with a velocity of 2.0 x \(10^{7}\) m/s.
(i) Write the complete nuclear equation of this decay reaction.
(ii) Write, but do not solve, a symbolic expression that could be used to calculate the mass released to energy in one single fermium-257 decay.
(iii) Assuming the fermium-257 is isolated and stationary, calculate the velocity of the Californium nucleus after the alpha particle is ejected.
Answer/Explanation
Ans:
Part (A)
Student A is correct that particles 2 and 3 have no net force acting on them while between the charged plates. This is evident in the fact that they travel in a straight line through the charged plate.
AND
Student A is correct in stating that particle 2 is neutral as it travels in a straight line through the magnetic field. Knowing that a charged particle will experience a force while passing through a magnetic field, particle 2 must be uncharged.
Student B is correct in stating that particle 3 must have a charge, since a moving charged particle will experience a force when passing through a magnetic field. Student B is also correct in stating that the path implies a negative charge. According to the right-hand rule for a charged particle in a magnetic field, the direction of force on a moving charged particle will be perpendicular to both the magnetic field and the direction of positive charge motion. Using this right-hand rule, we find a circular path directed upward for a positive charge, thus the downward curve of particle 3 suggests a negative charge.
AND
Student B is correct in stating that the bottom plate must be negatively charged. Since particle 3 is negative as previously stated, the negative charge will experience a downward force from the magnetic field. In order to maintain a straight line and thus no acceleration, particle 3 must experience an upward force to balance the downward force. This can be achieved only if the bottom plate repels the negative charge upward, indicating that the bottom plate is negatively charged.
Part (B)
Student A is incorrect in stating that particle 3 is neutral as evidenced by the fact that it experiences a force exerted by the magnetic field and arcs downward as soon as it leaves the electric field between the charged plates.
Student C is incorrect in stating that particle 1 is positively charged. Although the direction of the magnetic force on a positive charge would be upward, the bottom plate is negative and the top plate is positive. Thus, the electric force on particle 1 is downward. It is not possible to tell from the given information what the charge of particle 1 is. We can only say for sure that particle 1 has a net charge.
Part (C)
Part (D)
Substituting in the equation for velocity from above: 
Part (E)
Example: The mass can be found by dividing the charge by the charge to mass ratio (q/m). However, we know that particles come only in multiples of the electron charge. So, dividing the multiple of the electron charge by the charge to mass ratio will give us a set of possible masses for particle 3.
Part (F)
Question
The figure above shows a cross section of a cathode ray tube. An electron in the tube initially moves horizontally in the plane of the cross section at a speed of 2.0 x 107 meters per second. The electron is deflected
upward by a magnetic field that has a field strength of 6.0 x 10-4 tesla.
a. What is the direction of the magnetic field?
b. Determine the magnitude of the magnetic force acting on the electron.
c. Determine the radius of curvature of the path followed by the electron while it is in the magnetic field.
An electric field is later established in the same region as the magnetic field such that the electron now passes through the magnetic and electric fields without deflection.
d. Determine the magnitude of the electric field.
e. What is the direction of the electric field?
Answer/Explanation
Ans:
a) Using LHR for the electron, force up, velocity right, the B field points out of the page.
b) Fb = qvB = (1.6×10–19) (2×107) (6×10–4) = 1.9×10–15 N
c) Fnet(C) = mv2/r … qvB = mv2/ r … r = mv / Be … r = (9.11×10–31)(2×107) / (6×10–4)(1.6×10–19) = 0.19 m
d) ) Need Fe = Fb Eq = qvB E = vB E = (2×107)(6×10–4) = 12000 N/C
e) The E field must provide an electric force downwards on the negative charge to counteract the upwards B field. For a negative charge, this would require an upwards E field.