This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 27 Here’s 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 loop’s 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 ˆ x + B y ˆ y + B z ˆ 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 . Let’s 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 . Let’s now compute the individual magnetic forces on each segment, starting at the upperleft corner, and proceeding counterclockwise: F L = I ( L y ˆ y ) × B = IL y (ˆ y × B ) = IL y ˆ y × ( B x ˆ x + B y ˆ y + B z ˆ z ) = IL y [ B x (ˆ y × ˆ x ) + B y (ˆ y × ˆ y ) + B z (ˆ y × ˆ z )] = IL y ( B x ˆ z + B z ˆ x ) = IL y ( B z ˆ x B x ˆ z ) F D = I ( L x ˆ x ) × B = IL x (ˆ x × B ) = IL x ˆ x × ( B x ˆ x + B y ˆ y + B z ˆ z ) = IL x [ B x (ˆ x × ˆ x ) + B y (ˆ x × ˆ y ) + B z (ˆ x × ˆ z )] = IL x ( B y ˆ z B z ˆ y ) = IL x ( B z ˆ y + B y ˆ z ) 1 F R = I ( L y ˆ y ) × B = IL y (ˆ y × B ) = F L = IL y ( B z ˆ x B x ˆ z ) F U = I ( L x ˆ x ) × B = IL x (ˆ x × B ) = F D = IL x ( B z ˆ y + B y ˆ 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 (ˆ y × B ) + IL x (ˆ x × B ) + IL y (ˆ y × B ) IL x (ˆ x × B ) = 0 We’ve 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 ~μ : ~μ = ( I )(area) ˆ z = IL x L y ˆ 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 ˆ zdirection. This is the normal to the plane. We can now evaluate the torque on the loop. This is given by ~ τ = ~μ × B = IL x L y (ˆ z × B ) If we write B as B = B x ˆ x + B y ˆ y + B z ˆ z then the torque is explicitly given by ~ τ = IL x L y (ˆ z × B ) = IL x L y [ˆ z × ( B x ˆ x + B y ˆ y + B z ˆ z )] = IL x L y [ B x (ˆ z × ˆ x ) + B y (ˆ z × ˆ y ) + B z (ˆ z × ˆ z )] = IL x L y [ B x ˆ y B y ˆ x + B z (0)] = IL x L y ( B y ˆ x + B x ˆ y )...
View
Full Document
 Winter '07
 Kusenko
 Current, Magnetism, Force, Magnetic Field, loop, Clockwise, Brz − Blz

Click to edit the document details