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Figure 9.1.8 (a) A bar magnet moving toward a current loop. (b) Determination of the direction of induced current by considering the magnetic force between the bar magnet and the loop 9.2.Motional EMF Consider a conducting bar of length l moving through a uniform magnetic field which points into the page, as shown in Figure 9.2.1. Particles with charge q 0 inside experience a magnetic force ??=??× ?⃗ which tends to push them upward, leaving negative charges on the lower end. Figure 9.2.1 A conducting bar moving through a uniform magnetic field
104 The separation of charge gives rise to anelectric field ?⃗ inside the bar, which in turn produces a downward electric force ??= ??⃗ . At equilibrium where the two forces cancel, we have qvB qE, or E vB. Between the two ends of the conductor, there exists a potential difference given by Vab Va Vb El Blv (9.2.1) Since arises from the motion of the conductor, this potential difference is called the motional emf. In general, motional emf around a closed conducting loop can be written as 𝜀 = ∮(?× ?⃗ ) ∙ ?? (9.2.2) Where ?? is a differential length element.Now suppose the conducting bar moves through a region of uniform magnetic field ?⃗ = ??̂(pointing into the page) by sliding along two frictionless conducting rails that are at a distance l apart and connected together by a resistor with resistance R, as shown in Figure 9.2.2. Figure 9.2.2 A conducting bar sliding along two conducting rails Let an external force ????be applied so that the conductor moves to the right with a constant velocity ?= ??̂. The magnetic flux through the closed loop formed by the bar and the rails is given byB BA Blx (9.2.3) Thus, according to Faraday’s law, the induced emf is𝜀 = −?Φ???= −???(???) = −??????= −???(9.2.4) where dx / dt v is simply the speed of the bar. The corresponding induced current is? =|𝜀|?=????(9.2.5)
105 and its direction is counterclockwise, according to Lenz’s law. The equivalent circuit diagram is shown in Figure 9.2.3: Figure 10.2.3 Equivalent circuit diagram for the moving bar The magnetic force experienced by the bar as it moves to the right is ??= ?(??̂) × (−??̂) = −????̂ = − (?2?2??) ?̂(10.2.6) which is in the opposite direction of ?. For the bar to move at a constant velocity, the net force acting on it must be zero. This means that the external agent must supply a force ????= −??= + (?2?2??)?̂(9.2.7) The power delivered by ????is equal to the power dissipated in the resistor: ? = ????∙ ?= ????? = (?2?2??)? =(???)2?=𝜀2?= ?2?(9.2.8) as required by energy conservation.From the analysis above, in order forthe bar to move at a constant speed, an external agent must constantly supply a force ????. What happens if at t 0, the speed of the rod is v0, and the external agent stops pushing? In this case, the bar will slow down because of the magnetic force directed to the left. From Newton’s second law, wehave B2l2v dv FB Ror ma m dt (9.2.9)???=?2?2???? = −??𝜏(9.2.10)
106 0 where mR / B2l 2. Upon integration, we obtain v(t) v et /(9.2.11) Thus, we see that the speed decreases exponentially in the absence of an external agent doing work. In principle, the bar never stops moving. However, one may verify that the total distance traveled is finite.
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