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capa10 - Solution Derivations for Capa#10 1 A 1000-turn...

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Solution Derivations for Capa #10 1) A 1000-turn loop (radius = 0 . 038 m ) of wire is connected to a (25 Ω) resistor as shown in the figure. A magnetic field is directed perpendicular to the plane of the loop. The field points into the paper and has a magnitude that varies in time as B = gt , where g = 0 . 25 T/s . Neglect the resistance of the wire. What is the magnitude of the potential difference between points a and b ? N = Given r = Given (Radius) R = Given (Resistance) B = B ( t ) Given For a wire with multiple loops, the induced EMF is given by E = - N dt The magnetic flux φ is given by φ = B · dA which reduces to φ = BA . This is since magnetic field is always parallel to the area vector; thus, the dot product reduces to a simple multiplication. Also, the magnetic field is only changing in magnitude, not direction. Since the differential is with respect to the area, the fact that B is changing does not affect the integral and it becomes BA . Plugging back in, E = - N d dt ( BA ) = - NA dB ( t ) dt CAPA is looking for the magnitude of the answer. 2) What is the electrical energy dissipated in the resistor in 39 s ? t = Given To find energy, we must first find the power of the circuit. This is found by P = V 2 R = E 2 R where E is the induced EMF in problem 1. To find the energy, remember that power is defined to be energy per time, or the rate at which energy is used. Thus, energy is power times time. energy = P * t = E 2 R t 1
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3) A conducting rod is pulled horizontally with constant force F = 4 . 20 N along a set of rails separated by d = 0 . 420 m . A uniform magnetic field B = 0 . 500 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v = 3 . 70 m/s . Calculate the induced emf around the loop in the figure, due to Faraday’s Law and the changing flux. Assign clockwise to be the positive direction for emf. F = Given d = Given B = Given v = Given We know that E = - dt and φ = B · dA = B dA since the two vectors are parallel. Since B is constant, this again reduces to BA . This time the differential is changing. However, since everything else in the integrand was constant, dA is A regardless whether A is changing. But this value of A is the area at that instant the integral was taken. Since we are not concerned about the area at any particular instant, this does not matter. Plugging this back in, E = - d dt ( BA ) = - B dA dt since B is constant. The area of the loop at any time is given by length times width, or A = d * x . However, the length changes with time at a rate given by v . Since the velocity is the change in length per time, the length at time t is given by x = vt . Thus, we know that A = dvt . E = - B dA dt = - B d dt ( dvt ) = - Bdv Depending on which direction was assigned to be a positive current and which way the rod was moving, a negative sign might be in the v . When in doubt, use the sign given by the right hand rule.
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