Problem 5: A circular arc of wire with radius R and extent from 01 to 02 carries a current I as shown in the diagram to the right. (You may assume that this arc of wire is part of a larger circuit that may be ignored for the purposes of this exercise.) You will calculate the magnetic field at the point P by evaluating a Biot-Savart integral of the form B-/些 Ho Idš x î 4т R Part (a) Enter an expression for the current element I ds. Expression : I ds = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), ɛo, Ho, T, 01, 02, î, j, k, d0, dr, I, p, R Part (b) With the origin of your coordinate system placed at the observation point P, enter an expression for the unit vector î. Expression : r = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), £o, Ho, T, 01, 02, î, j, k, d0, dr, I, p, R Part (c) With the origin of your coordinate system placed at the observation point P, which integral provides a correct evaluation of the magnetic field at P? (Make no simplifications at this step.) MultipleChoice : cº IRde(- sin(@) i +cos(0) } ) × (- cos(0) i -sin(@) } ) 1) B R IRde(- sin(@) i +cos(8) )x(- cos(0) i -sin(8) } 2) B = R2 IRde(sin(0) i +cos(0) 3 ) × (- cos(@) i -sin(@) j 3) В - R IRde(- sin(0) i +cos(0) )x (- cos(0) i -sin(0) ) Ho 4) В R B = IRde (- sin(0) î +cos(0) j ) × (- cos(0) î – sin(0) 3 ) IRde(- sin(8) i +cos (0) ĵ ) × (- cos(@) i -sin(8) } ) 5) 47 6) IRde(- sin(@) i +cos(0) 3 )×(cos(@) i -sin(8) j) Ho 7) B= IRde(- sin(@) i +cos(0) 3 )× (cos(0) i +sin(0) 3 ) Ho 8) B = So, R Part (d) With the origin of your coordinate system placed at the observation point P, which partially simplified integral provides a correct evaluation of the magnetic field at P? MultipleChoice : 1) B Sa, cos²(0)de k 4mR B dri Jo, do k 2) Je, So (sin (0) – cos²(0)) do î: 3) 47R 4rR Je, sin? (0)de k So, 2 sin(@) cos(0)de k 4) B 5) B %3D 47R 4TR Je, de i Hol cos (0)de k 6) B 7) B %3D 47R Je B = 4R Sa* sin² (0)do k 8) Part (e) Evaluate the integral and enter a simplified expression for the magnitude of the magnetic field, B, evaluated at P. Expression : B = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), ɛo, Ho, T, 01, 02, î, j, k, d0, dr, I, p, R Part (f) Enter an expression for the direction of the magnetic field, B/B, evaluated at P. Expression : B/B = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), €o, Ho, T, 0, 01, 02, î, j, k, dé, dr, I, R 一5 一 I||| I| |||| I|||
Problem 5: A circular arc of wire with radius R and extent from 01 to 02 carries a current I as shown in the diagram to the right. (You may assume that this arc of wire is part of a larger circuit that may be ignored for the purposes of this exercise.) You will calculate the magnetic field at the point P by evaluating a Biot-Savart integral of the form B-/些 Ho Idš x î 4т R Part (a) Enter an expression for the current element I ds. Expression : I ds = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), ɛo, Ho, T, 01, 02, î, j, k, d0, dr, I, p, R Part (b) With the origin of your coordinate system placed at the observation point P, enter an expression for the unit vector î. Expression : r = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), £o, Ho, T, 01, 02, î, j, k, d0, dr, I, p, R Part (c) With the origin of your coordinate system placed at the observation point P, which integral provides a correct evaluation of the magnetic field at P? (Make no simplifications at this step.) MultipleChoice : cº IRde(- sin(@) i +cos(0) } ) × (- cos(0) i -sin(@) } ) 1) B R IRde(- sin(@) i +cos(8) )x(- cos(0) i -sin(8) } 2) B = R2 IRde(sin(0) i +cos(0) 3 ) × (- cos(@) i -sin(@) j 3) В - R IRde(- sin(0) i +cos(0) )x (- cos(0) i -sin(0) ) Ho 4) В R B = IRde (- sin(0) î +cos(0) j ) × (- cos(0) î – sin(0) 3 ) IRde(- sin(8) i +cos (0) ĵ ) × (- cos(@) i -sin(8) } ) 5) 47 6) IRde(- sin(@) i +cos(0) 3 )×(cos(@) i -sin(8) j) Ho 7) B= IRde(- sin(@) i +cos(0) 3 )× (cos(0) i +sin(0) 3 ) Ho 8) B = So, R Part (d) With the origin of your coordinate system placed at the observation point P, which partially simplified integral provides a correct evaluation of the magnetic field at P? MultipleChoice : 1) B Sa, cos²(0)de k 4mR B dri Jo, do k 2) Je, So (sin (0) – cos²(0)) do î: 3) 47R 4rR Je, sin? (0)de k So, 2 sin(@) cos(0)de k 4) B 5) B %3D 47R 4TR Je, de i Hol cos (0)de k 6) B 7) B %3D 47R Je B = 4R Sa* sin² (0)do k 8) Part (e) Evaluate the integral and enter a simplified expression for the magnitude of the magnetic field, B, evaluated at P. Expression : B = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), ɛo, Ho, T, 01, 02, î, j, k, d0, dr, I, p, R Part (f) Enter an expression for the direction of the magnetic field, B/B, evaluated at P. Expression : B/B = Select from the variables below to write your expression. Note that all variables may not be required. cos(0), sin(0), €o, Ho, T, 0, 01, 02, î, j, k, dé, dr, I, R 一5 一 I||| I| |||| I|||
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