Solution to UniPhysics_II_HW #5

Solution to UniPhysics_II_HW #5 - Solution to - HW...

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Solution to -- HW assignment #5 Chapter 33: Magnetic field and force --------------------------------------------------------------------------------------------------------------- (wires are long) Solve: The magnetic field strength at point a is () 00 at a top bottom top bottom 7 0 at a 22 5 at a , out of page , into page 11 1 1 21 0 T m /A1 0 A 2 2.0 cm 4.0 2.0 cm 2.0 10 m 6.0 10 m 6.7 10 T, out of page II BBB dd I B B μμ ππ μ π −− ⎛⎞ =+ = + ⎜⎟ ⎝⎠ ⇒= = × ⇒=× rrr r × At points b and c, 4 at 2 , into page , into page 2.0 10 T, into page B =+= × r 5 at 3 , into page , out of page B = × r ---------------------------------------------------------------------------------------------------------------------------------- (wires are long) Solve: (a) The Biot-Savart law (Equation 33.6) for the magnetic field of a current segment s Δ r is 0 2 ˆ Is r B 4 r Δ × = r r where the unit vector points from current segment Δ s to the point, a distance r away, at which we want to evaluate the field. For the two linear segments of the wire, ˆ r s Δ r is in the same direction as so ˆ, r ˆ 0. sr Δ ×= r For the curved segment, s Δ r and are always perpendicular, so . ˆ r ˆ s rs Δ× =Δ r Thus 0 Is 2 4 B r Δ = Now we are ready to sum the magnetic field of all the segments at point P. For all segments on the arc, the distance to point P is r = R . The superposition of the fields is 0 4 I Bd s arc 44 L R RR μμθ == = where L = R θ is the length of the arc. (b) Substituting = 2 in the above expression loop center 2 42 B R R μπμ This is Equation 33.7, which is the magnetic field at the center of a 1-turn coil.
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---------------------------------------------------------------------------------------------------------------------------------- ( one can just use the formula ) 4 0 θ π μ r I B center = Visualize: Please refer to Figure P33.47. The distance from P to the inner arc is r 1 and the distance from P to the outer arc is r 2 . Solve: As given in Equation 33.6, the Biot-Savart law for a current carrying small segment s Δ r is 0 ˆ Is r B 2 r 4 Δ × = r r For the linear segm of the loop, B Δ s = 0 T because ents ˆ 0. sr Δ×= r Consider a segment s Δ r on length on the inner arc. Because s Δ r is perpendicular to the ˆ r vector, we have 2 00 BB 1 0 0 0 0 arc 1 22 11 1 1 1 1 2 44 4 4 4 4 Is I r I I d I I rr r r r r μμ ππ ΔΔΔ = = = = A similar expression applies fo == r arc 2 B . The right-hand rule indicates an out-of-page direction for B arc 2 and an into-page direction for B arc 1 . Thus, 0 , into page , out of page , into page 4 II I B r r 12 1 2 ⎛⎞ =+ = ⎜⎟ ⎝⎠ r The field strength is ( ) () 7 5 41 0 T m / A 5 . 0 A 7.9 10 T B × =− 4 0.010 m 0.020 m = × Thus B r = (7.9 × 10 T, into page). --- -------------------------------- –5 ---------------------------------------------------------------------------------- ------- -------------------------------- Visualize: Ampere’s integration paths are shown in the figure for the reg < r < R 1 , R 1 < r < R 2 , and R 2 r . < ions 0 m Solve: For the region 0 m < r < R 1 , 0 through Bds I ⋅= r r ú . Because the current inside the integration path is zero, B = 0 T. To find I through in the region R 1 < r < R 2 , we rrent density by the area inside the integration path that carries the current. Thus, multiply the cu I I through 1 21 rR RR
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where the current density is the first term. Because the magnetic field has the same magnitude at every point on the circular path of
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This note was uploaded on 04/21/2011 for the course PHY 3240 taught by Professor Moore during the Spring '11 term at W. Florida.

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Solution to UniPhysics_II_HW #5 - Solution to - HW...

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