HW8solutions

# HW8solutions - Physics 213 Q28.8 Homework 8 Solutions...

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Physics 213 Homework 8 Solutions Spring 2008 Q28.8 If the current is clockwise in both loops, the B field due to the outer loop will be everywhere downward within that loop. The inner loop is embedded in this downward B field and will therefore feel an outward force on each segment that will try to expand the inner loop . The B field due to the inner loop will be everywhere upward outside of the inner loop, so the outer loop (embedded in this upward field) will feel in inward force on each segment that will try to contract the outer loop . If the direction of current is reversed, the directions of the magnetic forces remain the same for each loop because both the v and B vectors (or, equivalently, l and B vectors) change direction, and there is therefore no net change to v ± ± B ²± (or l ± ± B ). So either direction of current flow, the loops are pulled toward each other — just like two parallel currents attracting each other. 28.60. IDENTIFY: Find the vector sum of the magnetic fields due to each wire. SET UP: For a long straight wire B = μ 0 I 2 ± r . The direction of ± B is given by the right-hand rule and is perpendicular to the line from the wire to the point where then field is calculated. EXECUTE: (a) The magnetic field vectors are shown in Figure 28.60a. (b) At a position on the x -axis B net = 2 0 I 2 ± r sin = 0 I ± x 2 + a 2 a x 2 + a 2 = 0 Ia ± ( x 2 + a 2 ) , in the positive x -direction. (c) The graph of B versus x / a is given in Figure 28.60b. (d) The magnetic field is a max at the origin, x = 0. (e) When x >> a , B ± 0 Ia ± x 2 . Figure 28.60 28.67. IDENTIFY: Find the vector sum of the fields due to each loop. SET UP: For a single loop B = 0 Ia 2 2( x 2 + a 2 ) 3/2 . Here we have two loops, each of N turns, and measuring the field along the x- axis from between them means that the x ” in the formula is different for each case: EXECUTE: Left coil: x ± x + a 2 ² B left = 0 NIa 2 2(( x + a 2 ) 2 + a 2 ) 32 . ; Right coil: x ± x ² a 2 ³ B right = 0 NIa 2 2(( x ² a 2 ) 2 + a 2 ) So, the total field at a point a distance x from the point between them is B = 0 NIa 2 2 1 (( x + a 2) 2 + a 2 ) + 1 (( x ± a 2) 2 + a 2 ) ² ³ ´ µ · B 1x =B 1 sin ± ±

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Physics 213 Homework 8 Solutions Spring 2008 (b) B versus x is graphed in Figure 28.67. Figure 28.67a is the total field and Figure 27.67b is the field from the right- hand coil. (c) At point P , x = 0 and B = μ 0 NIa 2 2 1 (( a 2) 2 + a 2 ) + 1 (( ± a 2) 2 + a 2 ) ² ³ ´ µ · = 0 NIa 2 (5 a 2 4) = 4 5 ² ³ ´ µ · 0 NI a (d) B = 4 5 ± ² ³ ´ µ 32 0 NI a = 4 5 ± ² ³ ´ µ 0 (300)(6.00 A) (0.080 m) = 0.0202 T. (e) dB dx = 0 NIa 2 2 ± 3( x + a 2 ) (( x + a 2) 2 + a 2 ) + ± x ± a ) (( x ± a 2) 2 + a 2 ) ² ³ ´ µ · . At x = 0 , dB dx x = 0 = 0 NIa 2 2 ± a 2) (( a 2) 2 + a 2 ) 52 + ± ± a ) (( ± a 2) 2 + a 2 ) ² ³ ´ µ · = 0 . d 2 B dx 2 = 0 NIa 2 2 ± 3 (( x + a 2) 2 + a 2 ) + 6( x + a 2) 2 (5 2) (( x + a 2 ) 2 + a 2 ) ² ³ ´ + ± 3 (( x ± a ) 2 + a 2 ) 5/2 + 6( x ± a 2) 2 (5 2) (( x ± a 2 ) 2 + a 2 ) 7/2 µ · At x = 0 , d 2 B dx 2 x = 0 = 0 NIa 2 2 ± 3 (( a 2) 2 + a 2 ) 5/2 + 6( a 2) 2 (5 2) (( a ) 2 + a 2 ) + ± 3 (( a 2) 2 + a 2 ) 5/2 + 6( ± a 2) 2 (5 2) (( a 2 ) 2 + a 2 ) ² ³ ´ µ · = 0.
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HW8solutions - Physics 213 Q28.8 Homework 8 Solutions...

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