1. Show that the torque on any steady current distribution in a uniform field B is (m x B).(Problem 6.2)NOTE: (PLEASE SOLVE BY EXPLAİNİNG İNTERMEDİATE STEPS İNANSWERS.)(ANSWER)Problem 6.2 dF=ldlB;dN=r-adF=Ir-(dl-B). Now (Prob.1.6):r(dlB)+dl→(B»r)+B→¬(rdl)=0.Butd[r→»(r→B)] =dr¬(r¬B)+r¬(dr¬B)(sinceBis constant), anddr=dl, sodl-(B »r)=r»(dl»B)d[r»(r»B)]. Hence2r-(dl-»B)=Dd[r(r¬B)]B→¬(r→dl).dN=12{d[r(r→B)]B→»(r»dl)}.)N=12|Hd[r¬(r→B)]B-»H(r»dl). But the first term is zero (Hd()=0), and the second integral is2a(Eg.1.107). SON=I(B-a)=m»B.gedinin

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Answered: 1. Show that the torque on any steady…

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1. Show that the torque on any steady current distribution in a uniform field B is (m x B). (Problem 6.2)