List of Necessary and Unnecessary EquationsPhysics 260MWF 10:05 – 11:00 Dr. WombleName:_______________________________THE MORE WORK SHOWN, THE MORE CREDIT GIVEN!Constants:k=o41 = 9 x 109 Nm2/C2o = 8.85 x 10-12 C2/ Nm20 = 4 x 10-7 Tm/Amass electron, me = 9.11 x 10-31 kgelectron charge, e = -1.6 x 10-19 CConversions:1 eV =1.6 x 10-19 J1 u = 1.66 x 10-27 kg1) An alpha particle (q= +2e, mass=4.0 u) travels in a circular path of radius 4.5 cm in a magnetic field with B = 1.2 T. Calculate:a) its speedb) its period of revolutionc) its kinetic energy in electron volts (eV)d) the potential difference through which it had to be accelerated to achieve this kinetic energy (Hint: KE = qV)Bext2) The arrows in the object below indicate the initial orientation of magnetic dipoles in the material of which the object is composed. An external magnetic field, Bext,which is directed toward the left ofthe page is turned on.a) (4pts) What is the behavior of the magnetic dipoles if the material is paramagnetic? b) (4pts) Sketch this behavior in the space provided below.c) (4pts) What is the behavior of the magnetic dipoles if the material is diamagnetic?d) (4pts) Sketch this behavior in the space provided below.e) Ferromagnetism resembles the behavior of one of these types of materials. i) (1pt) Which type of material? ii) (2pts) In what ways is ferromagnetism similar to this type of material?iii) (2pts) In what ways is ferromagnetism different from this type of material?f) (4pts)Briefly describe how a material becomes magnetized.3) A small circular loop of area 0.0002 m2 is placed in the plane of, and concentric with, a large circular loop of radius 1 m. The current in the loop is changed uniformly from 200 A to -200 A (a change in direction) in a time of 1 s. a) What is the magnitude of the magnetic field at the center of the small circular loop due to the current in large loop at t=0.0 s? (13 pts)b) What EMF is induced in the small loop at t = 0.5 s? (Since the inner loop is small assume the field B (calculated in part a) is uniform over the area of the small loop) (Hint: Assume the change in the B field is the average value of B field during the entire 1 second period. Also, the final value of B is the negative of the value calculated in part a) (12 pts)Radius = 1 marea = 0.0002 m2Initial current = 200 AFinal current = -200 Aabc4) (25 pts) The figure below shows a cross-section of a coaxial cable and gives its radii (a,b,c). The current in the inner conductor is i but the current in the outer conductor is 2i. However, both are uniformly distributed in the two conductors. Derive expressions for the magnetic field B(r) for the following regions in:Hint: The left hand side of Ampere’s Law for this geometry is )2( rBdsBa) r < ab) a < r < bc) b< r <cd) r > cList of Necessary and Unnecessary Equations1. F = 22141rqqo2. F=qE3. E = rˆ412rqo4. E = qF5.aE dE6.encoq7. E=o8. E=o29. E = ro210. V = qW11.Vf - Vi = fidsE12. V= rqo4113. V=niiioniirqV114114. Es= -sV15. i =  AJ d16. R= iV17.JE118. R =AL19. P= iV = i2R = RV220. Fc = rmv221. K = ½ mv222. L = mvr23. T =vr224. f=T125. F = q v  B26. F = i L  B = N i A28. dB = 2ˆ4ridors 29. B =rio230. B =rio431. B= 0 i n32.encid0sB33. Fba= ibLBa34. B  dA 035.E s d -ddtB36.B s  dddti0 0E0 enc  37. E=dtd-B38. id =dtd E039. E = i R40.aB