EDT 16 Electromagnetic induction. Solutions

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Stuart Barker
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1 EDT 16 Electromagnetic induction Solutions 1. Consider a strip of length dx at distance x as shown in the figure. Flux through this strip is dφ = BA = (B 0x) (a dx) = B 0a x dx. Integrating this from x = 0 to x = a, we get, Φ = B 0a 3 /2 a 2. We want flux through the surface to be maximum. So, center of the loop should be kept at a point where magnetic field is maximum i.e. where B 1x B 2x 2 is maximum. d dx (B1x B2x2 ) = 0 => B 1 x = 0 => x = B 1 So X coordinate of center should be at B 1 i.e. x 0 + a 2 = B 1. So x 0 = B 1 a Φ = B. A = A ( B.n ) = A (B. n )/ n where A is the numerical value of the area and n is the unit vector along the normal. Φ = 7 x 10-5 Wb 4. Φ = B x (Area projected perpendicular to magnetic field) = πr 2 B. 5. Magnetic flux through a closed surface will be zero. 6. Apply Lenz law. 7. Flux through the surface is zero as it is parallel to the field. Therefore, no EMF will be induced in the loop. 8. When magnet is moving towards magnet, current will be clockwise. When it is moving away from coil, current will be anti-clockwise. 9. According to Lenz law, current in the coil will be set so that is oppose the motion of magnet. 10. At t = 3, slope of graph is positive. Therefore, flux in the upward direction is increasing. To reduce this effect, current will be in clockwise direction. x dx

2 11. Induced EMF will be maximum when the rate of change of flux will be maximum i.e. when dφ is maximum. As area is constant, dφ will be maximum when 9 3t 2 6t 3t 2 6t db is maximum i.e. when 3t2 6t is maximum. 3 Note the modulus sign. It is must to have modulus as we are not interested in direction of EMF and EMF of -5V is same as that of 5V magnitude wise The graph of 3t 2 6t is shown in the figure. In the range t = 0 to t = 2 maximum value occurs at t = 1s. 12. In the range t = 0 to t = 3 maximum value occurs at t = 3s. Note that though t = 3 is not local maxima, value of function is maximum at that point. 13. Assume that angle made by velocity with X-axis is θ. Then horizontal component of velocity is vcosθ while vertical component is vsinθ. If particle is situated at [h, k], then in time it will be at [h + (vcosθ), k + (vsinθ)]. Field at [h, k] was B i = 3h + 4k. Now at [h + (vcosθ), k + (vsinθ)] field is B f = 3[h + (vcosθ)] + 4[k + (vsinθ)] E = dφ = A db = A B f B i = A [3vcosθ + 4vsinθ] When we try to maximize the EMF by differentiating w.r.t. θ, we get that tanθ = 4/ Obvious. 15. Φ = B A = B π r 2 = πr 0 2 B e -2αt. E = dφ = 2απr02 Be -2αt. 16. Obvious. 17. The electric field developed in this case is not conservative and hence we do not have potentials defined. 18. Informative. 19. Area vector in this case can be taken as (100 cm 2 ) 1 (-3i + 4j ) 5 Other things are now obvious. 20. Magnetic field inside the toroid is given by μ 0NI 2πR. Hence flux is Φ = μ 0NI 0 t 2πR x a Note that we will use area of toroid and not the ring. This is because part of area enclosed by the ring has no flux while only that part which is inside has uniform magnetic field. EMF = dφ = μ 0NI 0 a 2πR

3 Projected length 21. Assume side of square is a. Then a = 10 cm. dφ = Badx = Bav = 0.5 mv 22. Obvious. 23. Consider a small element of length dr at distance r from the center. Velocity of this element is ωr. Hence induced emf dε = B ωr dr. Integrating this from 0 to L, we get, ε = BωL 2 /2 24. See the previous question. Use the same method. Just limits of integration will change from L/4 to 3L/ Use the Projected length concept or Effective length concept. That is consider this as rod with a length equal to the projection of rod perpendicular to its velocity. In the given case it is H. So EMF will be ε = vbl = vbh. 26. Again projected length is R. So EMF will be ε = vrb. 27. Projected length is zero. EMF is zero. v L H 28. Vertical component of the Earth s magnetic field is Bsinδ. Force on the conductor will be F = BLv sinδ. 29. Current generated in the wire will be I = ε R = BvL R. Weight of the wire will be balanced by the force due to magnetic field. Hence, mg = BIL = B 2 L 2 v/r => v = mgr/b 2 L Use Flemming s rule. 31. εab = BL(v/2) εpq = B(2L)(v/2) Hence ratio is 1 : Informative. 33. U = E V = B2 2μ => E = B 2 2μ R μ 0 V = B2 AL 2μ R μ 0 = B2 πr 2 L 2μ R μ Obvious. 35. E = ½ Li 2 Hence, P = de di = Li = 0.02W 36. Let the number of turn in the two solenoids be 2n and n respectively. M = μ 0πn 1n 2r 12 L => π 2 x 10-6 = 4π x 10-7 x π x 2n 2 x (0.05) 2 x 0.2 n = Informative. 38. L1 2 L2 = (n 1) x l 1 = 1 : 1. n 2 l Self-inductance is property of Solenoid and is independent of current or potential differences.

4 40. As d R, Magnetic field due to coil of radius R at the center of other coil will be Flux through the other coil Φ = BA = μ 0R 2 Iπr 2 Hence L = μ 0πr 2 R 2 2d ε = M di = 20 mv. 42. Φ = LI = 3.6 Wb. 43. Informative. 2x 3 B = μ 0R 2 I 2x ε = NABω = NAB(2πf) 45. If number of turns is increased by factor of k while keeping the total length of the wire same, then perimeter must decrease by factor of k. This means that length as well as breah of the coil will decrease by factor of k. So area will decrease by factor of k 2. As N increase by k times and A decreased by k 2 times, net EMF generated will decrease by factor of k.

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Demo: Solenoid and Magnet. Topics. Chapter 22 Electromagnetic Induction. EMF Induced in a Moving Conductor

Demo: Solenoid and Magnet. Topics. Chapter 22 Electromagnetic Induction. EMF Induced in a Moving Conductor Topics Chapter 22 Electromagnetic Induction EMF Induced in a Moving Conductor Magnetic Flux EMF Induced in a Moving Conductor Demo: Solenoid and Magnet v 1 EMF Induced in a Moving Conductor q Work done