Determine the magnetic force on a very thin wire, whichfollows a simicircular cu

Question

Determine the magnetic force on a very thin wire, whichfollows a simicircular curve of radius R = 4m, and lies in theupper half of the x-y plane with its center at the origen. There is a constant current i = 2 A flowing counterclockwise,starting upward from the end of the of the wire on the positivex-axis and ending downward at the end of the negativex-axis. The wire is in a uniform magnetic field B = 3 T. Thefield B is in a direction parallel to the z-axis. (Solution check: 48 N in the y direction). Pleaseshow step-by-step solution. Thank you.

Explanation / Answer

The current in the half-loop goes in the negative x directionand also goes in the positive y direction and then the negative ydirection. . The B field in the z direction applies force to current in xdirection and current in y direction. But here's the trick: .          Since thecurrent goes in the positive and then negative y directions, theforce on the current in the y direction cancels out. The totalforce on the "y current" is zero! . This means that the total force on the half-circle is only onthe current that moves in the x direction. The total length of the"x current is just   2R (i.e. the distance the currentmoves in the x direction). So now... .      force on current in B field = L I B =   2 R I B = 2 * 4 * 2 *3 =   48 N . And the direction? This is found using the right hand rule.The direction of force is the direction of current (cross) Bfield, which in your case is    negative x (cross) z .   The result of this is  positive y. . So total force on the half-circle is    48 N inthe +y direction

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