B 5 m 0 i p r a b i d s 5 b 1 2 p r 2 2 i r r 2 b 5 m

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B 5 m 0 I p r . a B i D s 5 B 1 2 p r 2 . 2 I . r . R 2 : B 5 m 0 I 2 p r . a B i D s 5 B 1 2 p r 2 . R 1 , r , R 2 : R 3 I I R 2 R 1 B i 5 B . B 5 0. B 1 2 p r 2 5 0 I encl 5 0, r . R 3 , B 5 m 0 I 2 p r . B 1 2 p r 2 5 m 0 I . I encl 5 I , R 1 , r , R 2 , a B i D s 5 B 1 2 p r 2 . B S B 5 m 0 I 2 p r . B 1 2 p r 2 5 m 0 I I encl 5 I , a B i D s 5 B a D s 5 B 1 2 p r 2 . B S a B i D s 5 m 0 I encl . I 5 2.80 3 10 2 5 T # m 4 p 3 10 2 7 T # m / A 5 22.3 A. m 0 I 5 2.80 3 10 2 5 T # m I encl m 0 I encl , B i D s Magnetic Field and Magnetic Forces 20-21
20.78. Set Up: The toroidal solenoid is sketched in Figure 20.78. A few representative turns of wire are shown in the sketch. These turns carry current into the page at the inner edge of the solenoid and carry current out of the page at the outer edge of the solenoid. Apply Ampere’s law to a circular path of radius r. By symmetry the magnetic field is constant on the path and tangent to it, so Figure 20.78 Solve: (a) No current passes through the path so the enclosed current is zero. Ampere’s law gives and Each wire passes through the path twice, once carrying current into the page and once carrying current out of the page. Therefore, the net current through the path is zero. Ampere’s law gives and (b) Each of the N turns carries current through the area enclosed by the path and the enclosed current is Ampere’s law gives and 20.79. Set Up: with The magnetic field in the material is a factor of greater than it is in vacuum. Solve: (a) (b) Reflect: The magnetic field inside the paramagnetic material is larger than the external field, but only slightly larger. 20.80. Set Up: Solve: (a) (b) m 5 K m m 0 5 1 1.0015 21 4 p 3 10 2 7 T # m / A 2 5 1.259 3 10 2 6 T # m / A K 5 1.5023 T 1.5000 T 5 1.0015. m 5 K m m 0 K m 5 B inside / B outside . B inside 5 K m B external 5 1 1.00026 21 1.3500 T 2 5 1.3504 T m 5 1 1.00026 21 4 p 3 10 2 7 T # m / A 2 5 1.257 3 10 2 6 T # m / A K m m 0 5 4 p 3 10 2 7 T # m / A. m 5 K m m 0 , B 5 m 0 IN 2 p r . B 1 2 p r 2 5 m 0 NI NI . a B i D s 5 B 1 2 p r 2 . a , r , b : B 5 0. B 1 2 p r 2 5 0 a B i D s 5 B 1 2 p r 2 . r . b : B 5 0. B 1 2 p r 2 5 0 a B i D s 5 B 1 2 p r 2 . r , a : I I I I I I r B i 5 B . 20-22 Chapter 20
20.81. Set Up: Use to calculate With downward, and The direction of the force is given by the right-hand rule and the magnitude is given by The charge of the ball is Solve: The directions of and are shown in Figure 20.81, where the downward direction is into the page. The right-hand rule (rhr) direction is north, but since the charge is negative the force is opposite the rhr direction and is to the south. At the bottom of the shaft the speed of the ball is so Figure 20.81 Reflect: The magnetic force is much less than the gravity force on the ball. 20.82. Set Up: The magnetic force must be upward and equal to mg. The direction of is determined by the direction of I in the circuit. with where V is the battery voltage. Solve: (a) The forces are shown in Figure 20.82. The current I in the bar must be to the right to produce upward.
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