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There is no magnetic force on the portions of the wire outside the magnetic field region.

Example: a wire carrying current I consists of a semicircle of radius R and two horizontal straight portions each of length L. It is in a region of constant magnetic field as shown. What is the net magnetic force on the wire?. y. . . . . . . . . B. . . . . . . .

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There is no magnetic force on the portions of the wire outside the magnetic field region.

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  1. Example: a wire carrying current I consists of a semicircle of radius R and two horizontal straight portions each of length L. It is in a region of constant magnetic field as shown. What is the net magnetic force on the wire? y         B         x R         I L L                 There is no magnetic force on the portions of the wire outside the magnetic field region.

  2. First look at the two straight sections.         F1 F2 B         R         I L L         L  B, so y         x

  3. Next look at the semicircular section.         dF F1 d F2 B ds         R Calculate the infinitesimal force dF on an infinitesimal ds of current-carrying wire.          I L L         y         x Why did I call that angle  instead of ? ds subtends the angle from  to +d. Because we usually use  for the angle in the cross product. The infinitesimal force is ds  B, so Arc length Finally,

  4. dFy Calculate the y- component of F.       dF F1 d F2 B ds      R       I L L         y         x Interesting—just the force on a straight horizontal wire of length 2R.

  5. Does symmetry give you Fx immediately?       dF F1 d F2 B ds      dFx R Or, you can calculate the x component of F.       I L L         y         x Sometimes-Useful Homework Hint Symmetry is your friend.

  6. Fy Total force:         dF F1 F2 B ds         R          I L L         y         x We probably should write the force in vector form. Possible homework hint: how would the result differ if the magnetic field were directed along the +x direction? If you have difficulty visualizing the direction of the force using the right hand rule, pick a ds along each different segment of the wire, express it in unit vector notation, and calculate the cross product.

  7. Example: a semicircular closed loop of radius R carries current I. It is in a region of constant magnetic field as shown. What is the net magnetic force on the loop of wire? FC y         B         x R         I                 We calculated the force on the semicircular part in the previous example (current is flowing in the same direction there as before).

  8. FC Next look at the straight section.         B         R         I         L  B, and L=2R so y         FS x Fs is directed in the –y direction (right hand rule). The net force on the closed loop is zero! This is true in general for closed loops in a uniform magnetic field.

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