PHYS 272 Study Guide - Final Guide: Surface 2, Wavelength, Inductor
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If the ole(cid:374)oid is (cid:373)o(cid:448)ed to(cid:449)ards the loop, the (cid:272)urre(cid:374)t (cid:894)(cid:271)(cid:455) le(cid:374)z"s la(cid:449)(cid:895) is i(cid:374)du(cid:272)ed a(cid:374)d the dire(cid:272)tio(cid:374) is counter clockwise. The total resistance of the loop is r= 0. 280 . Calculated the force f required to pull the loop from the field (to the right) with a constant velocity of v=3. 4m/s. The external force must then balance this magnetic force: use farada(cid:455)"s la(cid:449) to (cid:272)o(cid:373)pute the i(cid:374)du(cid:272)ed emf and current, thus the power dissipated in the resistance of the loop. This energy must be supplied by the external force on the loop. =(cid:882). (cid:889)(cid:889)(cid:885)(cid:887) (cid:1839)(cid:1832) (cid:1866)(cid:1856)(cid:1873)(cid:1855)(cid:1857)(cid:1856) (cid:1855)(cid:1873)(cid:1870)(cid:1870)(cid:1857)(cid:1866)(cid:1872)= (cid:1844)(cid:1857)(cid:1871)(cid:1871)(cid:1872)(cid:1866)(cid:1855)(cid:1857)=(cid:884). (cid:889)(cid:888) (cid:1827) (cid:1832)(cid:1867)(cid:1870)(cid:1855)(cid:1857)=(cid:4666) (cid:1828)(cid:4667)=(cid:884). (cid:889)(cid:888) (cid:882). (cid:889)(cid:887) (cid:882). (cid:888)(cid:887) =(cid:884). (cid:883)(cid:885) (cid:1849: this problem is about determining the direction of the electric field induced by a changing magnetic field. This first step is to obtain the direction of the magnetic field due to the current. The(cid:374), tr(cid:455) to use the a(cid:374)alog(cid:455) of farada(cid:455)"s la(cid:449), i. e. (cid:1856)=(cid:1856)(cid:3003) (cid:1856)(cid:1872) (cid:449)ith a(cid:373)pere"s la(cid:449),
