Magnetism #2 - Questions and Answers

Question 1 The magnetic field B in acertain region is 0.128, and its direction is that of the z-axis in the figure. Z y b 30.0 cm 40.0 cm e a C 30.0 cm f X d 50.0 cm Part A What is the magnetic flux across the surface abcd in the figure? Part B What is the magnetic flux across the surface befc ? Part C What is the magnetic flux across the surface aefd? Part D What is the net flux through all five surfaces that enclose the shaded volume? Answer 1 α) (abcd) = B. A = 0 b) B(befc) = B. A = −(0.128T)(0.300 m) (0.300 m) = −0.0115Wb C) фв(aefd) BA = BAcos 3 (0.128T)(0.500 m)(0.300 m) = +0.0115Wb d) The net flux through the rest of the surface is zero since they are parallel to the x-axis so the total flux is the sum of all parts above, which is zero. Answer 2 Step 1 When there is a change in magnetic flux then there is induced emf in the coil and the direction of the induced emf is such that it opposes the change and its direction is given by Lenz law. The magnetic flux (4) can be given by, $ = BA where, B is the magnetic field A is the area Step 2 a) The magnetic flux across the surface 'abcd' is, $ = BA cos $ = 0.128T × (0.4m x 0.3m) × cos 90° since area vector is in negative x direction while magnetic field is in positive z direction. $ = 0 b) The magnitude flux across the surface 'befc' is, $ = BA cos $ = 0.128T × (0.3m x 0.3m) × cos 180° since area vector is in negative z direction while magnetic field is in positive z direction. = -0.0115Wb Step 3 c) The magnetic flux across the surface 'aefd' is, The area of surface 'aefd' is, A = 0.5 x 0.3 = 0.15m² The angle can be calculated as, 30cm cos 0 = 50cm $ = BA cos $ = 0.128T × (0.15) × $ = 0.0115Wb Step 4 d) The net flux through all five surfaces that enclose the shaded volume is given as, ф = abcd + befc + aefd + abe + $cdf ф = 0 – 0.0115 +0.0115 +0+0 $ = 0 Question 2 An airplane pilot sets a compass course due west and maintainsan airspeed of 220 km/h. After flying for 0.500 h, she find sherself over a town 120 km west and 20 km south of her starting point. a) Find the wind velocity (magnitude and direction). b) If the wind velocity of 40 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 km/h. The answers are: a) 44.7 km/h, 26.6 deg. west of south b) 10.5 deg. north of west Answer 1 If there is no wind then theplane would be at a distance d=(220 km/h)(0.5h)=110km west of the starting point. if the wind is blown then then the plane is 120km-110km=10km west and 20km south in the time 0.5h. Velocity of the plane in west direction=(10km)/0.5h=20km/h velocity of the plane in south direction=(20km)/0.5h=40km/h Now the wind velocity is 2 √(20km)² + (40km)² = 44.7km h the direction is 40k h h 0 = tan(10) = 63° 20k south of west h b) If the wind velocity is 40km/h then the direction is 0 = sin 1( 40km = 10.5° 220k h north of west Answer 2 The velocity of the plane relative to earth is given by VP/E == 0.5 -120-20) = [−240 – 40j] km/h According to relative motion the velocity of the air relative to the earth is given by VA/E = VP/E - VP/A = −240 – 40 – (-2202) = [-20 – 40j] km/h The magnitude v_{A/E} is given by UA/E = √(40)2 + (20)2 = 44.7 km/h the angle 0₁ is given by 01 = arctan(- -40 () = 63.4° But the angle is in 3rd quadratic so add 180 01=180+63.4 = 243.4° b)In this case the captain want to move with 220 km/h toward west relative to earth and wind blow in 40 km/h toward south relative to earth too, so the vectors are given by: VP/E = [-220] km/h VA/E = [-40j] km/h The relative velocity VP/A is given by VP/A = VP/E-VA/E = -220 – (-40)) = [-220 + 40j] km/h The magnitude UP/A is given by UP/A = √(220)2 + (40)2 = 223.6 km/h The angle 01 is also given by 02 = arctan(220) = -10.3° The angle 02 is in 2nd quadratic so add 180 to get it with respect to +x -axis 02 = 180 - 10.3 = 169.7° Or we can say 10.3° north of west Question 3 Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 13 m into 35 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 72 kg, find the magnitudes of the impulse on him from the water. Answer 1 I solved the question: Given that The height of belly-flop, h = 13 m mass of Henri LaMothe, m = 72 kg The potential energy of him at a height h is, PE = mgh Let the initial velocity when he reaches the water surface = v 1 2 The kinetic energy at the water surface, KE = =mv From the law of conservation of energy 2 -mv² = mgh 1 2 vo = √2gh 2 = 2(9.8 m/s²)(13m) = 15.96 m/s when he reaches the bottom of the water, final velocity v₁ = 0 m/s Therefore, the change in momentum is given by Ap = m(v -vo) = (72 kg)(0-15.96 m/s) =-1149.12 kg.m/s Therefore, the magnitude of the impulse on him from water is Ap=-1149.12 kg.m/s Answer 2 Given that 1149.12 N-s The height of belly-flop, h = 13 m mass of Henri LaMothe, m = 72 kg The potential energy of him at a height h is PE = mgh Let the initial velocity when he reaches the water surface = vo The kinetic energy at the water surface, KE = mv From the law of conservation of energy mv = mgh √2gh = √2(9.8 m/s²)(13 m) = 15.96 m/s when he reaches the bottom of the water, final velocity vf = 0m/s Therefore, the change in momentum is given by Ap = m(vf - vo) = (72 kg)(0-15.96 m/s) = 1149.12 kg.m/s Therefore, the magnitude of the impulse on him from water is |Ap| = | -1149.12 kg-m/s = 1149.12 N-s