Magnetism and Moving Charges AP Physics 2 FRQ – Exam Style Questions etc.
Magnetism and Moving Charges AP Physics 2 FRQ
Unit 12: Magnetism and Electromagnetic Induction
Weightage : 15–18%
Exam Style Practice Questions , Magnetism and Moving Charges AP Physics 2 FRQ
Question
The long, straight wire shown in Figure 1 above is in the plane of the page and carries a current I. Point P is also in the plane of the page and is a perpendicular distance d from the wire. Gravitational effects are negligible.
a. With reference to the coordinate system in Figure 1, what is the direction of the magnetic field at point P due to the current in the wire?
A particle of mass m and positive charge a is initially moving parallel to the wire with a speed vo when it is at point P. as shown in Figure 2 below.
b. With reference to the coordinate system in Figure 2, what is the direction of the magnetic force acting on the particle at point P ?
c. Determine the magnitude of the magnetic force acting on the particle at point P in terms of the given quantities and fundamental constants.
d. An electric field is applied that causes the net force on the particle to be zero at point P.
i. With reference to the coordinate system in Figure 2, what is the direction of the electric field at point P that could accomplish this?
ii. Determine the magnitude of the electric field in terms of the given quantities and fundamental constants.
Answer/Explanation
Ans:
a) Based on the RHR, the B field is directed into the page on the – z axis.
b) Based on the RHR, the force is directed down on the – y axis.
c) First determine the B field at point A from the wire. B = µoI / (2πd)
Then the force on the particle is given by Fb =qvB = qvoµoI / (2πd)
d) Since the magnetic force is directed down, the electric force would have to act upwards to cancel. Since the charge is positive, the E field would also have to point upwards in the +y direction
e) Fe = Fb Eq = qvB E = vB E = voµoI / (2πd)
Question
A charged particle is projected from point P with velocity v at a right angle to a uniform magnetic field directed out of the plane of the page as shown. The particle moves along a circle of radius R.
(a) On the diagram, draw a vector representing the magnetic force acting on the particle at point P.
(b) Determine the sign of the charge of the particle. Explain your reasoning.
(c) Explain why the magnetic field does no work on the particle as it moves in its circular path.
(d) A second, identically charged particle is projected at position P with a speed 2v in a direction opposite that of the first particle. On the diagram, draw the path followed by this particle. The drawn path should include a calculation of the radius of curature in terms of the original radius R.
Answer/Explanation
Ans:
(a) Vector pointing toward center of circle.
(b) Negative. In order to apply the hand rule, the fingers of the left hand point in the direction of the particle’s velocity, and bend out of the plane of the page, leaving the left-hand thumb pointing toward the center of the circle, creating the centripetal force allowing the particle to move in a circular path.
(c) No work is done because the magnetic force is always perpendicular to the velocity of the particle.
(d) See diagram at right. Radius of the new circle is 2R, since \(mv^2/R=qvB\), therefore \(R=mv/qB\). 