ohw18-solutions

# ohw18-solutions - howard(cah3459 ohw18 turner(56705 This...

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howard (cah3459) – ohw18 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points A closed circuit consists oF two semicircles oF radii 77 cm and 38 cm that are connected by straight segments, as shown in the fgure. A current oF I = 3 A ±ows around this circuit in the clockwise direction. The permeability oF Free space is 4 π × 10 - 7 Tm / A. O 38 cm 77 cm ˆ ı ˆ ˆ k ˆ k is upward From the paper I = 3 A I What is the direction oF the magnetic feld B at point O . 1. / B ± / B ± = +ˆ 2. None oF these are true‘. 3. / B ± / B ± = +ˆ ı 4. / B ± / B ± = - ˆ 5. / B ± / B ± =+ ˆ k 6. / B ± / B ± = - ˆ ı 7. / B ± / B ± = - ˆ k correct Explanation: IF you curve the fngers on your right hand in the direction oF the current, the magnetic feld points along your thumb. 002 (part 2 of 2) 10.0 points ²ind the magnitude oF the magnetic feld at point at point O . Correct answer: 3 . 7042 μ T. Explanation: Let : r 1 = 77 cm = 0 . 77 m , r 2 = 38 cm = 0 . 38 m , I = 3 A , and μ 0 =4 π × 10 - 7 / A . The magnetic feld due to a circular loop at its center is B = μ 0 I 2 R . Because the loop is only halF oF a current loop, the magnetic feld due to the arc 1 is / B 1 = - μ 0 I 4 r 1 ˆ k, and the magnetic feld due to the arc 2 is / B 2 = - μ 0 I 4 r 2 ˆ k. Thus the total magnetic feld is / B = / B 1 + / B 2 = - μ 0 I 4 ± 1 r 1 + 1 r 2 ² ˆ k = - (4 π × 10 - 7 / A) (3 A) 4 × ± 1 0 . 77 m + 1 0 . 38 m ² ˆ k ± 10 6 μ T 1T ² =( - 3 . 7042 μ T) ˆ k ± / B ± = 1 . 003 10.0 points The closed loop shown in the fgure carries a current oF 20 A in the counterclockwise direc- tion. The radius oF the outer arc is 60 cm , that oF the inner arc is 40 cm . Both arcs ex- tent an angle oF 60 . The permeability oF Free space is 4 π × 10 - 7 / A.

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howard (cah3459) – ohw18 – turner – (56705) 2 40 cm C B 60 cm D A O 60 y z Find the component of the magnetic ±eld at point O along the x axis. Correct answer: - 1 . 74533 × 10 - 6 T. Explanation: Let : r i = 40 cm = 0 . 4m , r o = 60 cm = 0 . 6m , I = 20 A , and μ 0 =4 π × 10 - 7 Tm / A . The magnetic ±eld due to a circular loop at its center is B = μ 0 I 2 R . Because the loop is only one-sixth of a current loop, the magnetic ±eld due to the inner arc is / B i = - μ 0 I 12 r i ˆ ı, and the magnetic ±eld due to the outter arc is / B o = μ 0 I 12 r o ˆ ı. Thus the total magnetic ±eld is / B = / B i + / B o = μ 0 I 12 ± 1 r o - 1 r i ² ˆ ı = (4 π × 10 - 7 / A) (20 A) 12 × ± 1 0 . - 1 0 . ² ˆ ı = ( - 1 . 74533 × 10 - 6 T ) ˆ ı . 004 10.0 points Given: Three conductors (same length) have the shape of an equilateral triangle (whose sides are of length . 3 ), a ring (whose radius is of length r ), a square (whose sides are of length . 4 ). These conductors’ lengths are all equal (perimeters are 3 . 3 =2 π r . 4 ). All conductors carry the same counter- clockwise current I as shown in the ±gure below. . 3 I P . 4 I P r I P Select the correct comparison for the mag- nitude of the magnetic ±eld at the center points P of the current loops shown above.
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ohw18-solutions - howard(cah3459 ohw18 turner(56705 This...

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