# hw17 - homework 17 LEE BENJAMIN Due 4:00 am Question 1 chap...

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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 θ B = 0 . 0262 T B = 0 . 0262 T i = 1 . 59 A y 0 . 398 m 0 . 398 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). μ θ φ B mg x z Let θ be the angle the plane of the loop makes with the z axis as shown. Then the angle the coil’s magnetic moment μ makes with the z axis is φ = 90 θ ; e.g. , sin φ = cos θ and tan φ = cot θ . 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 τ g = vectorr × vector F = parenleftbigg 2 cos φ parenrightbigg m g , 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 τ m = N B I A sin φ . At equilibrium, τ m = τ g N B I ℓ 2 sin φ = m g ( cos φ ) 2 .

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homework 17 – LEE, BENJAMIN – Due: Oct 13 2007, 4:00 am 2 Thus tan φ = m g 2 N B I ℓ = (0 . 327 kg) (9 . 8) 2 (19) (0 . 0262 T) (1 . 59 A) (0 . 398 m) = 5 . 08638 . Since tan φ = tan(90 θ ) = cot θ , the angle the loop makes with the z axis at equilibrium is θ = cot 1 (5 . 08638) = 11 . 1227 .
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