Magnetism and Current- Carrying Wires AP Physics 2 FRQ – Exam Style Questions etc.
Magnetism and Current- Carrying Wires AP Physics 2 FRQ
Unit 12: Magnetism and Electromagnetic Induction
Weightage : 15–18%
Exam Style Practice Questions , Magnetism and Current- Carrying Wires AP Physics 2 FRQ
Question
a. Two long parallel wires that are a distance 2a apart carry equal currents I into the plane of the page as shown above.
i. Determine the resultant magnetic field intensity at the point O midway between the wires.
ii. Develop an expression for the resultant magnetic field intensity at the point N. which is a vertical distance y
above point O. On the diagram above indicate the direction of the resultant magnetic field at point N.
Answer/Explanation
Ans:
a) The field from a wire is given by B = µoI / (2πR), with R and I equal for both wires at point O. Based on the RHR
for the current wires, the right wire makes a field down and the left wire makes a field up so cancel to zero.
b) 
Based on the RHR, the resultant fields from each wire are directed as shown. Since the distance to each wire is the same, the resultant B field will simply be twice the x component of one of the wire’s B fields.
The distance to point N is \(\sqrt{a^{2} + y^{2}}\) so the total field at that location from a single wire is \( B = \frac{\mu _{0}I}{2\pi {\sqrt{a^{2} + y^{2}}}}\)
The x component of that field is given by B cos θ, where cos θ can be replaced with cos θ = a/h = y / \(\sqrt{a^{2} + y^{2}}\)
Giving Bnet = 2 B cos θ = 2 B y / \(\sqrt{a^{2} + y^{2}}\) = \(\frac{2y\mu _{0}I}{2\pi {\sqrt{a^{2} + y^{2}}}\sqrt{a^{2} + y^{2}}}\) = \(\frac{y\mu _{0}I}{\pi \left ( a^{2} + y^{2} \right )}\)
Question
In a uniform magnetic field B directed vertically downward. a metal bar of mass m is released from rest and slides without friction down a track inclined at an angle θ, as shown. The electrical resistance of the bar between its two points of
contact with the track is R. The track has negligible resistance. The width of the track is L.
a. Show on the diagram the direction of the current in the sliding bar.
b. Denoting by v the instantaneous speed with which the bar is sliding down the incline, determine an expression for the magnitude of the current in the bar.
c. Determine an expression for the force exerted on the bar by the magnetic field and state the direction of that force.
d. Determine an expression for the terminal velocity of the sliding bar.
Answer/Explanation
Ans:
a) The rod forms a loop with the upper part of the rails. Based on Lenz law, as the rod slides down, the perpendicular component of B increases the flux in the loop and current flows to create a field in the outward
normal direction to the rails to counteract the flux change. Using the RHR solenoid, the current would flow towards the right side of the bar pictured.
b) The induced emf in the bar is given by ε = Blv ⊥ . The perpendicular B is given with B cos θ (simiar to Fgx)
so the induced emf is BLv cos θ. With V = IR … This gives, IR = BLv cos θ and the current is I = BLv cos θ / R
c) F = BIL ⊥, again using B cos θ we have Fb = B cos θ I L, now sub in I from above. Fb = B cos θ(BLv cos θ / R)L
Fb = B2L2v cos2 θ / R. Based on the RHR for the bar, the force acts up the inclined rails parallel to them.
d) The terminal velocity will occur when Fnet(x) = 0 with x being the direction along the rails. This gives:
Fgx = Fb mg sin θ = B2L2v cos2 θ / R v = mg sin θ R / B2L2 cos2 θ