1Bch27 - Chapter 27 Heres an important point to keep in...

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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 xy-plane. The current travels counterclockwise as seen from above the plane. The loops dimensions are l x in the x-direction, and l y in the y-direction. 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 upper-left 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 xy-plane, the direction perpendicular to this is the z-direction. 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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1Bch27 - Chapter 27 Heres an important point to keep in...

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