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

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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)
NOTE: (PLEASE SOLVE BY EXPLAİNİNG İNTERMEDİATE STEPS İN
ANSWERS.)
(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)]. Hence
2r-(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.ged
inin
Transcribed Image Text: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 İN ANSWERS.) (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)]. Hence 2r-(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.ged inin
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