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Unformatted text preview: homework 17 LEE, BENJAMIN Due: Oct 13 2007, 4:00 am 1 Question 1, chap 30, sect 3. part 1 of 2 10 points A long piece of wire with a mass of 0 . 327 kg and a total length of 30 . 248 m is used to make a square coil with a side of 0 . 398 m. The coil is hinged along a horizontal side (the y axis), carries a 1 . 59 A current, and is placed in a magnetic field with a magnitude of 0 . 0262 T in the vertical direction along the z axis as shown in the figure below. The acceleration of gravity is 9 . 8 . z x B =0 . 0262T B =0 . 0262T i = 1 . 59 A y . 398 m . 3 9 8 m Determine the angle that the plane of the coil makes with the z axis when the coil is in equilibrium. Correct answer: 11 . 1227 (tolerance 1 %). Explanation: Let : L = 30 . 248 m , = 0 . 398 m , m = 0 . 327 kg , i = 1 . 59 A , and B = 0 . 0262 T . Look down the positive y-axis at the coil (that is, from the right-hand side of the origi- nal figure). B mg x z Let be the angle the plane of the loop makes with the z axis as shown. Then the angle the coils magnetic moment makes with the z axis is = 90 ; e.g. , sin = cos and tan = cot . The number of turns in the loop is N = L circumference = 30 . 248 m 4 (0 . 398 m) = 19 . The torque about the z-axis due to gravity is g = vectorr vector F = parenleftbigg 2 cos parenrightbigg mg , where is the length of each side of the square loop. This gravitational torque tends to ro- tate the loop clockwise. The torque due to the magnetic force tends to rotate the loop counterclockwise about the z-axis and has magnitude m = N B I A sin . At equilibrium, m = g N B I 2 sin = mg ( cos ) 2 . homework 17 LEE, BENJAMIN Due: Oct 13 2007, 4:00 am 2 Thus tan = mg 2 N B I = (0 . 327 kg) (9 . 8) 2 (19) (0 . 0262 T) (1 . 59 A) (0 . 398 m) = 5 . 08638 . Since tan = tan(90 ) = cot , the angle the loop makes with the z axis at equilibrium is = cot 1 (5 . 08638) = 11 . 1227 ....
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