f5ch30 - 800 CHAPTER 30 AMPERES LAW CHAPTER 30 Answer to...

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800 CHAPTER 30 AMPERE’S LAW C HAPTER 30 Answer to Checkpoint Questions . ( a ), ( c ), ( b ) . b , c , a . d , tie of a and c , then b . ( d ), ( a ), tie of ( b ) and ( c ) (zero) Answer to Questions . ( c ), ( d ), then ( a ) and ( b ) tie . ( a ) into; ( b ) greater . and . ( c ), ( a ), ( b ) . ( a ), ( b ), ( c ) (see Eq. - ) . ( c ), ( a ), ( b ) . ( b ), ( d ), ( a ) (zero) . ( a ) , , ; ( b ) less than . ( a ) , + x ; , y ; ( b y ; , + x . a , b , c . outward . a , tie of b and d , then c . c and d tie, then b , a . b , a , d , c (zero) . ( d ), then tie of ( a ) and ( e ), then ( b ), ( c ) . ( a ) opposite ; ( b ) and opposite ; ( c ) and opposite and ; ( d ) and opposite , , and . (scalar product is zero)
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CHAPTER 30 AMPERE’S LAW 801 . ( a ) a ; ( b d Solutions to Exercises & Problems E Solve i from Eq. - : i = rB = ( : m)( : T) T m/A = : A : E Use Eq. - : B = i r = ( T m/A)( A) ( : m = ) : T : E ( a ) The magnitude of the magnetic eld due to the current in the wire, at a point a distance r from the wire, is given by B = i r : Put r ft = : m. Then B = ( T m/A)( A) ( : m) : T = : T : ( b ) This is about one-sixth the magnitude of the Earth's eld. It will a ect the compass reading. E The current i due to the electron ow is i = ne = ( : = s)( : C) = : A. Thus B = i r = ( T m/A)( : A) ( : m) : T :
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802 CHAPTER 30 AMPERE’S LAW E Use B ( x; y; z ) = ( = ) i s r =r , where s = s j and r = x i + y j + z k . Thus B ( x; y; z ) = i s j ( x i + y j + z k ) ( x + y + z ) = = i s ( z i x k ) ( x + y s + z ) = : ( a ) B ( ; ; : m) = ( T m/A)( : A)( : m)( : m) i [ + + ( : m) ] = = ( : i ) T : ( b ) B ( ; : m ; ), since x = z = . ( c ) B ( : m ; : m ; ) = ( T m/A)( : A)( : m)( : m) k [( : m) + ( : m) ] = = ( : k ) T : ( d ) B ( : m ; : m ; ) = ( T m/A)( : A)( : m)( : m) k [( : m) + ( : m) ] = = ( : k ) T : E The points must be along a line parallel to the wire and a distance r from it, where r satis es B wire = i r = B ext ; or r = i B ext = ( : T m/A)( A) ( : T) : m : E ( a ) The eld due to the wire, at a point : cm from the wire, must be T and must be directed due south. Since B = i= r , i = rB = ( : m)( T) T m/A A :
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d x y current charge z CHAPTER 30 AMPERE’S LAW 803 ( b ) The current must be from west to east to produce a eld to the south at points below it. E Set up a coordinate system as shown to the right. The B - eld at the location of the charge q is B = i d ( k ) : Thus F q = q v B = i d ( k v ) : ( a ) Now v = v ( i ) so F q = iv d k ( i ) = iqv d ( j ) ; i.e., F q has a magnitude of iqv= d and is directed against the current direction. ( b ) Now the direction v is reversed so F q = iqv j = d . E The straight segment of the wire produces no magnetic elds at C . The eld from the two semi-circular loop cancel at C . So B C = . E Use the same coordinated system as in E. Then F e = ( e iv= d )( k v ), as obtained there. ( a ) Now v = v ( i ) so F e = iev j d = ( : T m/A)( A)( : C)( : m/s) j ( : m) = ( : N) j : ( c ) Now v = v k so F e / k v = : P ( a ) The straight segment of the wire produces no magnetic eld at C . ( b ) For the semicircular loop B C = Z loop j d s r j r = i Z loop r ds r = ( i )( r ) R = i R :
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ds r P 804 CHAPTER 30 AMPERE’S LAW B points into the page. ( c ) The same as ( b ), since the straight segments do not contribute to B . P Since sections AH and JD do not contribute to B C , we only need to consider the two arcs. For the smaller arc B C = Z i d s r r = i Z R ds R e = i e R ; where e is a unit vector pointing into the page. Similarly, for the larger arc B C = i e R : Thus B C = B C + B C = i ( R R ) e R R : P First nd the magnetic eld of a circular arc at its center. Let d s be an in nitesimal segment of the arc and r be the vector from the segment to the arc center. d s and r are perpendicular to each other, so the contribution of the segment to the eld at the center has magnitude dB = ids r :
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