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WileyPLUS c21 q.1.gif 684x471 pixels WileyPLUS: MywilleyPLUSI Help I Contact Us I Log Cutnell, Physics, 9e College Physics I and II (PHY 2053/20 ice Assignment Gradebook nent ent FULL SCREEN PRINTER VERSION BACK NEXT Chapter 21, Problem 38 GO A copper rod of length 0.96 m is tying on a frictionless table (see A copper rod of length 0.96 m is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of k 80 N/ the current in the copper rod that causes the m. A magnetic field with a strength of 0.20 T is oriented perpendicular to the surface of the table. (a) What must be the direction of springs to stretch? (b) If the current is 15 A, by how much does each spring stretch? (a) Direction (b) Number the tolerance is +/-2% Click if you would like to show work for this question: Units OpenSho·work Question Attempts: 0 of S used SAVE FOR LATERSUBHIT ANSWER SUBMIT ANSWER mapleNET

Explanation / Answer

magnetic force F on a rod is given by F = iLB sintheta

where i is curruent, B is magentic field, L is length , theta is the angle = 90 deg

also for spring constant k, F = kx

part A : direction of current will be perpendicular to the rod

part B :

x = F/2K as there are two springs connected in parallel

so

x = (15* 0.96 * 0.2)/(2* 80)

x = 1.8 cm

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