Question

4, A conductive bar of length L = 0.8 m is pulled to the left at a constant velocity of 1.8 m/s. A magnetic field of magnitude 0.06 T is directed out of the page. a) What is the induced voltage in the loop? (10 pts.) What is the magnitude and direction of the induced current, if the resistance of the circuit is 2.5 Ω? (5 pts.) b) As the induced current passes through the movable bar, it couples to the magnetic field, and the bar experiences an additional force due to this effect. What is the magnitude of this force? (10 pts.) c)

Answer 1

(a)

The orientation of bar is important, which is missing in the question. Lets assume that the bar is pulled perpendicular to its length. Then,

The orientation of bar is important, which is missing in the question. Lets assume that the bar is pulled perpendicular to its length. Then, E_emf = BLv = 0.06 times 0.8 times 1.8 = 0.0864 N Current flows downwards. Magnitude, I = E_emf/R = 0.0864/2.5 = 0.03456 Amp F = BIL = 0.06 times 0.03456 times 0.8 = 1.66 times 10^-3 N

(b)

Current flows downwards. Magnitude,

$I=E_{emf}/R=0.0864/2.5=0.03456 \textup{ Amp}$

(c)

$F=BIL=0.06\times0.03456\times0.8= 1.66\times10^{-3}\textup{ N}$