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Fall 2011 Homework 17 Solutions

# Fall 2011 Homework 17 Solutions - Physics 21 Fall 2011...

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Physics 21 Fall, 2011 Solution to HW-17 29-7 The current in the long, straight wire AB shown in the figure is upward and is increasing steadily at a rate di/dt . (a,b) At an instant when the current is i , what are the mag- nitude and direction of the field B at a distance r to the right of the wire? (c) What is the flux d Φ B through the narrow shaded strip? (d) What is the total flux through the loop? (e) What is the induced emf in the loop? (f) Evalu- ate the numerical value of the induced emf if a = 12 . 0 cm, b = 36 . 0 cm, L = 24 . 0 cm, and di/dt = 9 . 60 A/s. (a,b) B = μ 0 i ( t ) 2 πr into page (by right hand rule) (c) d Φ = B dA = μ 0 i ( t ) 2 πr Ldr (d) Integrate Φ = integraldisplay b a d Φ = μ 0 i ( t ) L 2 π integraldisplay b a dr r = μ 0 i ( t ) L 2 π ln b a (e) |E| = d Φ dt = μ 0 L 2 π di dt ln b a (f) (2 × 10 - 7 )(0 . 24m)(9 . 6) parenleftbigg ln 36 12 parenrightbigg = 5 . 06 × 10 - 7 V What will be the direction of the current? Counterclockwise. 29-8 A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure. The field is changing with time, according to B ( t ) = (1 . 4T) e - (0 . 057 s - 1 ) t . (a) Find the emf induced in the loop as a function of time (assume t is in seconds). (b) When is the induced emf equal to 1/20 of its initial value? (c) Find the direction of the current induced in the loop, as viewed from above the loop. (a) We can find the emf induced in the loop using contintegraldisplay E · d l = - d dt integraldisplay B · d A = - d Φ dt , where the integral of E · d l is the induced emf. Because the magnetic field makes a 60 degree angle with the loop, we can write integraldisplay B · d A = Φ = B ( t ) A sin(60) where A = πr 2 is the area of the loop. Because the area of the loop doesn’t change with time, d Φ /dt = πr 2 dB/dt , where d dt B ( t ) = (1 . 4)( - 0 . 057) e - 0 . 057 t Putting these pieces together results in: E = - πr 2 sin(60)(1 . 4)( - 0 . 057) e - 0 . 057 t = 0 . 122 e - 0 . 057 t We usually take E to be the magnitude of the emf, and don’t worry about the sign. The negative sign is related to the polarity of the voltage. (b) Now that we have an expression for the induced emf as a function of time, we can solve for t at a particular emf.

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