2017 USA Physics Olympiad Exam Part A 9 component directed tangent to a circle of radius d around the ball,so using polar coordinates is recommended. Solution This problem is not nearly as difficult as it looks. North 中 Consider the colored triangle above.The black side has length a.The angle between the green and black sides is o,so the length of the red side is asino and the length of the green side is a coso. The magnetic field strength from one magnetic pole a distance d away is given by 1 B=上mP The sum of the two fields has two components.The angular component is a measure of the "opening"of the triangle formed by the two vectors,and since the two vectors basically have the same length,we can use similar triangles to conclude asin。Bo d 户B B。=m8sm6=B,m d sin o. As expected,this component vanishes for=0. The radial component is given by the difference in the lengths of the two field vectors,or a+ m2x d2 d' where r=a coso is the length of the green side,so Ki Br =2Be" cosΦ. d That wasn't so bad,was it? d.If placed directly to the right and left of the ship compass,the iron balls can be located at a distance d to cancel out the error in the magnetic heading for any angle(s)where 60 is largest. Assuming that this is done,find the resulting expression for the combined deviation 60 due to the ship and the balls for the magnetic heading for all angles 0. Copyright C2017 American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 9 component directed tangent to a circle of radius d around the ball, so using polar coordinates is recommended. Solution This problem is not nearly as difficult as it looks. φ North a Consider the colored triangle above. The black side has length a. The angle between the green and black sides is φ, so the length of the red side is a sin φ and the length of the green side is a cos φ. The magnetic field strength from one magnetic pole a distance d away is given by B = ±m 1 d 2 The sum of the two fields has two components. The angular component is a measure of the “opening” of the triangle formed by the two vectors, and since the two vectors basically have the same length, we can use similar triangles to conclude a sin φ d ≈ Bφ B ⇒ Bφ = m a d 3 sin φ = Be mKi d 3 sin φ. As expected, this component vanishes for φ = 0. The radial component is given by the difference in the lengths of the two field vectors, or Br = m  1 d 2 − 1 (d + x) 2  = m d 2  1 − 1 (1 + x/d) 2  ≈ m d 2 2x d , where x = a cos φ is the length of the green side, so Br = 2Be mKi d 3 cos φ. That wasn’t so bad, was it? d. If placed directly to the right and left of the ship compass, the iron balls can be located at a distance d to cancel out the error in the magnetic heading for any angle(s) where δθ is largest. Assuming that this is done, find the resulting expression for the combined deviation δθ due to the ship and the balls for the magnetic heading for all angles θ. Copyright c 2017 American Association of Physics Teachers