Chapter 30 - CHAPTER Magnetic Induction 30 1* A uniform...

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CHAPTER 30 Magnetic Induction 1* A uniform magnetic field of magnitude 2000 G is parallel to the x axis. A square coil of side 5 cm has a single turn and makes an angle θ with the z axis as shown in Figure 30-28. Find the magnetic flux through the coil when ( a ) = 0 ° , ( b ) = 30 ° , ( c ) = 60 ° , and ( d ) = 90 ° . ( a ), ( b ), ( c ), ( d ) φ m = BA cos ( a ) m = 2 × 10 –1 × 25 × 10 –4 Wb = 5 × 10 –4 Wb = 0.5 mWb; ( b ) m = 0.433 mWb; ( c ) m = 0.25 mWb; ( d ) m = 0 2 A circular coil has 25 turns and a radius of 5 cm. It is at the equator, where the earth's magnetic field is 0.7 G north. Find the magnetic flux through the coil when its plane is ( a ) horizontal, ( b ) vertical with its axis pointing north, ( c ) vertical with its axis pointing east, and ( d ) vertical with its axis making an angle of 30 ° with north. ( a ), ( b ), ( c ) Use Equ. 30-3 ( a ) m = [(25 × 7 × 10 –5 × π × 25 × 10 –4 ) cos 90 ° ] Wb = 0 ( b ) m = (1.37 × 10 –5 cos 0 ° ) Wb = 1.37 × 10 –5 Wb ( c ) m = (1.37 × 10 –5 cos 30 ° ) Wb = 1.19 × 10 –5 Wb 3 A magnetic field of 1.2 T is perpendicular to a square coil of 14 turns. The length of each side of the coil is 5 cm. ( a ) Find the magnetic flux through the coil. ( b ) Find the magnetic flux through the coil if the magnetic field makes an angle of 60 ° with the normal to the plane of the coil. ( a ), ( b ) Use Equ. 30-3 ( a ) m = (14 × 25 × 10 –4 × 1.2) Wb = 0.042 Wb ( b ) m = (0.042 cos 60 ° ) Wb = 0.021 Wb 4 A circular coil of radius 3.0 cm has its plane perpendicular to a magnetic field of 400 G. ( a ) What is the magnetic flux through the coil if the coil has 75 turns? ( b ) How many turns must the coil have for the flux to be 0.015 Wb? ( a ), ( b ) Use Equ. 30-3 ( a ) m = (75 × × 9 × 10 –4 × 4 × 10 –2 ) Wb = 8.48 mWb ( b ) N = 75(15/8.48) = 133 5* A uniform magnetic field B is perpendicular to the base of a hemisphere of radius R . Calculate the magnetic flux through the spherical surface of the hemisphere. Note that m through the base must also penetrate the spherical surface. Thus, m = R 2 B . 6 ∙∙ Find the magnetic flux through a solenoid of length 25 cm, radius 1 cm, and 400 turns that carries a current
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Chapter 30 Magnetic Induction of 3 A. Use Equs. 29-9 and 30-3; φ m = µ 0 N 2 AI / L m = 7.58 × 10 –4 Wb 7 ∙∙ Work Problem 6 for an 800-turn solenoid of length 30 cm, and radius 2 cm, carrying a current of 2 A. Use Equs. 29-9 and 30-3; m = 0 N 2 AI / L m = 6.74 × 10 –3 Wb 8 ∙∙ A circular coil of 15 turns of radius 4 cm is in a uniform magnetic field of 4000 G in the positive x direction. Find the flux through the coil when the unit vector perpendicular to the plane of the coil is ( a ) n = i , ( b ) n = j , ( c ) n = ( i + j )/ 2 , ( d ) n = k , and ( e ) n = 0.6 i + 0.8 j . ( a ), ( b ), ( c ), ( d ), ( e ) Use Equ. 30-1 with B = 0.4 T i ( a ) NA = 0.0704 m 2 ; m = 0.0302 Wb. ( b ) m = 0 ( c ) Wb 0.0213 = Wb 2 / 0.0302 = m . ( d ) m = 0 ( e ) m = 0.0169 Wb 9* ∙∙ A solenoid has n turns per unit length, radius R 1 , and carries a current I . ( a ) A large circular loop of radius R 2 > R 1 and N turns encircles the solenoid at a point far away from the ends of the solenoid. Find the magnetic flux through the loop. ( b ) A small circular loop of N turns and radius R 3 < R 1 is completely inside the solenoid, far from its ends, with its axis parallel to that of the solenoid. Find the magnetic flux through this small loop.
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This homework help was uploaded on 04/09/2008 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter 30 - CHAPTER Magnetic Induction 30 1* A uniform...

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