Question

A. The drawing shows a bar magnet falling through a metal ring. In part a the ring is solid all the way around, but in part b it has been cut through.

S S N N (a) (b)

1. Explain why the motion of the magnet in part a is retarded when the magnet is above the ring and below the ring as well.

2. Draw any induced currents that appear in the ring.

3. Explain why the motion of the magnet is unaffected by the ring in part 2.

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B.

A magnetic field of magnitude B = 0.20 T is reduced to zero in a time interval of Δt = 0.10 s, thereby creating an induced current in a loop of wire. Which one or more of the following choices would cause the same induced current to appear in the same loop of wire?

B = 0.40 T and Δt = 0.20 s

B = 0.30 T and Δt = 0.10 s

B = 0.30 T and Δt = 0.30 s

B = 0.10 T and Δt = 0.050 s

B = 0.50 T and Δt = 0.40 s

Answer 1

© In A the ring, the magnetic field changes (increases). an oppose the paut When the magnet approaches associated with the ring so emf is induced in the sing. According to Lenz's law, the magnetic field and due to the induced emf will always cause ie., the magnetic field due to bau magnet. so the two each other and it retards. Whos the bas magnet goes below field associated with ring decreases . So an emf is induced. The magnetic field due to the induced emt opposes the motion of bau magnet clause). 30 then also it is retarded. fields oppose the magnetic 2 15 G [ Here we use lena's law and right hand thumb rule I CS Scanned with CamScanner

peth ③ in paut la , the ring is cut. 3o the induced current donot have a closed to flowe. so the magnet is unaffected B Let À be the corea of the loop resistance. dB 0.2 -0 -0.2T dt =0 Induced emf, E = do 03 - BA at d CBA) loop & R'be the - S = at E = A A dB at Induced current, I 1 I = A dB R at 0.2 - Axor . I- 2A R 0.4 0 dB = 0'4T ; dt =0.25 I= - 02 2A/R possible I CS Scanned with CamScanner

dB = 0.31 ; dt oilS .: I = A x 0.3 - 31/2 > not possible dB = 0.31 ; dt = 0.35 :: I - A х 0.3 0:3 - AR > not possible. dB -0.17 ; dt = 0.05 S .; I A R O'OS 2A/R → possible. dB -0.5 T ; dt =0.45 ... I 0.5 А R Х -25 À → Not poesible options © ¢ are possible. CS Scanned with CamScanner