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Unformatted text preview: Chapter 27 Heres an important point to keep in mind regarding magnetic forces. Consider a closed loop of any shape in a uniform magnetic field, carrying a current I . The magnetic force on the loop is zero . This can be shown mathematically as follows: F = Z loop Id l B = Z loop B Id l = B I Z loop d l = B ( I )(0) = 0 However, the magnetic torque on the loop is generally nonzero. Problem 1: A rectangular loop carrying current I lies in the xyplane. The current travels counterclockwise as seen from above the plane. The loops dimensions are l x in the xdirection, and l y in the ydirection. A uniform magnetic field B fills the entire region, where B = B x x + B y y + B z z 1: Evaluate explicitly the magnetic force on each of the four segments of the loop. Show that the total magnetic force on the loop is zero. 2: What is the magnetic torque on the loop? Solution: We first consider part 1 . Lets label the left vertical segment as segment L , the right vertical segment as segment R , the upper horizontal segment as segment U , and the lower horizontal segment as segment D . Lets now compute the individual magnetic forces on each segment, starting at the upperleft corner, and proceeding counterclockwise: F L = I ( L y y ) B = IL y ( y B ) = IL y y ( B x x + B y y + B z z ) = IL y [ B x ( y x ) + B y ( y y ) + B z ( y z )] = IL y ( B x z + B z x ) = IL y ( B z x B x z ) F D = I ( L x x ) B = IL x ( x B ) = IL x x ( B x x + B y y + B z z ) = IL x [ B x ( x x ) + B y ( x y ) + B z ( x z )] = IL x ( B y z B z y ) = IL x ( B z y + B y z ) 1 F R = I ( L y y ) B = IL y ( y B ) = F L = IL y ( B z x B x z ) F U = I ( L x x ) B = IL x ( x B ) = F D = IL x ( B z y + B y z ) The total magnetic force on the loop is the sum of these individual forces: F = F L + F D + F R + F U = IL y ( y B ) + IL x ( x B ) + IL y ( y B ) IL x ( x B ) = 0 Weve just shown that for a closed rectangular loop in a uniform magnetic field, the magnetic force is zero. We now consider part 2 . To find the torque on the loop, we must first find the magnetic dipole moment ~ : ~ = ( I )(area) z = IL x L y z Notice that the direction of the dipole moment is perpendicular to the plane of the loop. Because the loop lies in the xyplane, the direction perpendicular to this is the zdirection. This is the normal to the plane. We can now evaluate the torque on the loop. This is given by ~ = ~ B = IL x L y ( z B ) If we write B as B = B x x + B y y + B z z then the torque is explicitly given by ~ = IL x L y ( z B ) = IL x L y [ z ( B x x + B y y + B z z )] = IL x L y [ B x ( z x ) + B y ( z y ) + B z ( z z )] = IL x L y [ B x y B y x + B z (0)] = IL x L y ( B y x + B x y )...
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 Winter '07
 Kusenko
 Current, Magnetism, Force

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